Vì hàm số  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacYgacaGGVbGaai4zamaaBaaaleaacaWGJbaabeaakiaadIha % aaa!3CE3! y = {\log _c}x\)nghịch biến nên 0 < c < 1, các hàm số \( y = {a^x},\,y = {b^x}\)đồng biến nên a >  1; b > 1 nên c là số nhỏ nhất trong ba số.

Đường thẳng x = 1 cắt hai hàm số \( y = {a^x}, y ={b^x}\)tại các điểm có tung độ lần lượt là a và b, dễ thấy a > b. Vậy c < b < a

Đặt \(t= 2^x , t > 0\) ta được phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaCa % aaleqabaGaaGOmaaaakiabgkHiTiaaisdacaWG0bGaey4kaSIaaG4m % aiabg2da9iaaicdacqGHuhY2daWabaabaeqabaGaamiDaiabg2da9i % aaigdaaeaacaWG0bGaeyypa0JaaG4maaaacaGLBbaaaaa!46B6! {t^2} - 4t + 3 = 0 \Leftrightarrow \left[ \begin{array}{l} t = 1\\ t = 3 \end{array} \right.\)

Với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmamaaCa % aaleqabaGaamiEaaaakiabg2da9iaaigdacqGHuhY2caWG4bGaeyyp % a0JaaGimaaaa!3EBD! {2^x} = 1 \Leftrightarrow x = 0\) và với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmamaaCa % aaleqabaGaamiEaaaakiabg2da9iaaiodacqGHuhY2caWG4bGaeyyp % a0JaciiBaiaac+gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOGaaG4maa % aa!4284! {2^x} = 3 \Leftrightarrow x = {\log _2}3\)

Dạng đồ thị hình bên là đồ thị hàm đa thức bậc 3; \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadggacaWG4bWaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaamOy % aiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWGJbGaamiEai % abgUcaRiaadsgaaaa!431A! y = a{x^3} + b{x^2} + cx + d\)  có hệ số a > 0.

Do đó, chỉ có đồ thị ở đáp án A là thỏa mãn.

Vì phương trình \(f(x) = 2018\) có ba nghiệm phân biệt nên đồ thị hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaaGymaaqaaiaadAgadaqadaqaaiaadIhaaiaawIca % caGLPaaacqGHsislcaaIYaGaaGimaiaaigdacaaI4aaaaaaa!4014! y = \frac{1}{{f\left( x \right) - 2018}}\) có ba đường tiệm cận đứng.

Mặt khác, ta có :

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcqGHRaWkcqGHEisP % aeqaaOGaamyEaaaa!3F42! \mathop {\lim }\limits_{x \to + \infty } y\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaC % beaeaaciGGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcqGHRaWk % cqGHEisPaeqaaOWaaSaaaeaacaaIXaaabaGaamOzamaabmaabaGaam % iEaaGaayjkaiaawMcaaiabgkHiTiaaikdacaaIWaGaaGymaiaaiIda % aaaaaa!4766! = \mathop {\lim }\limits_{x \to + \infty } \frac{1}{{f\left( x \right) - 2018}}\)\(= - \frac{1}{{2019}}\)

nên đường thẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Jaey % OeI0YaaSaaaeaacaaIXaaabaGaaGOmaiaaicdacaaIXaGaaGyoaaaa % aaa!3BA6! = - \frac{1}{{2019}}\) là đường tiệm cận ngang của đồ thị hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaaGymaaqaaiaadAgadaqadaqaaiaadIhaaiaawIca % caGLPaaacqGHsislcaaIYaGaaGimaiaaigdacaaI4aaaaaaa!4014! y = \frac{1}{{f\left( x \right) - 2018}}\)

Và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcqGHsislcqGHEisP % aeqaaOGaamyEaaaa!3F4D! \mathop {\lim }\limits_{x \to - \infty } y\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaC % beaeaaciGGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcqGHsisl % cqGHEisPaeqaaOWaaSaaaeaacaaIXaaabaGaamOzamaabmaabaGaam % iEaaGaayjkaiaawMcaaiabgkHiTiaaikdacaaIWaGaaGymaiaaiIda % aaaaaa!4771! = \mathop {\lim }\limits_{x \to - \infty } \frac{1}{{f\left( x \right) - 2018}}\)=0 nên đường thẳng y = 0 là đường tiệm cận ngang của đồ thị hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaaGymaaqaaiaadAgadaqadaqaaiaadIhaaiaawIca % caGLPaaacqGHsislcaaIYaGaaGimaiaaigdacaaI4aaaaaaa!4014! y = \frac{1}{{f\left( x \right) - 2018}}\)

Vậy \(k + l = 5\)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGtbGaamytaaqaaiaadofacaWGbbaaaiabg2da9iaadUgaaaa!3B42! \frac{{SM}}{{SA}} = k\) với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4AaiabgI % GiopaadmaabaGaaGimaiaacUdacaaIXaaacaGLBbGaayzxaaaaaa!3C8E! k \in \left[ {0;1} \right]\)

Xét tam giác SAB có MN // AB nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGnbGaamOtaaqaaiaadgeacaWGcbaaaiabg2da9maalaaabaGaam4u % aiaad2eaaeaacaWGtbGaamyqaaaacqGH9aqpcaWGRbaaaa!3F8A! \frac{{MN}}{{AB}} = \frac{{SM}}{{SA}} = k\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % ytaiaad6eacqGH9aqpcaWGRbGaaiOlaiaadgeacaWGcbaaaa!3E2B! \Rightarrow MN = k.AB\)

Xét tam giác SAD có MQ // AD nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGnbGaamyuaaqaaiaadgeacaWGebaaaiabg2da9maalaaabaGaam4u % aiaad2eaaeaacaWGtbGaamyqaaaacqGH9aqpcaWGRbaaaa!3F8F! \frac{{MQ}}{{AD}} = \frac{{SM}}{{SA}} = k\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % ytaiaadgfacqGH9aqpcaWGRbGaaiOlaiaadgeacaWGebaaaa!3E30! \Rightarrow MQ = k.AD\)

Kẻ đường cao SH của hình chóp. Xét tam giác SAH có:

MM' // SH nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGnbGabmytayaafaaabaGaam4uaiaadIeaaaGaeyypa0ZaaSaaaeaa % caWGbbGaamytaaqaaiaadofacaWGbbaaaaaa!3DA5! \frac{{MM'}}{{SH}} = \frac{{AM}}{{SA}}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaWGtbGaamyqaiabgkHiTiaadofacaWGnbaabaGaam4uaiaa % dgeaaaGaeyypa0JaaGymaiabgkHiTmaalaaabaGaam4uaiaad2eaae % aacaWGtbGaamyqaaaacqGH9aqpcaaIXaGaeyOeI0Iaam4Aaaaa!4681! = \frac{{SA - SM}}{{SA}} = 1 - \frac{{SM}}{{SA}} = 1 - k\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % ytaiqad2eagaqbaiabg2da9maabmaabaGaaGymaiabgkHiTiaadUga % aiaawIcacaGLPaaacaGGUaGaam4uaiaadIeaaaa!417F! \Rightarrow MM' = \left( {1 - k} \right).SH\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGnbGaamOtaiaadcfacaWGrbGaaiOlaiqad2eagaqbaiqa % d6eagaqbaiqadcfagaqbaiqadgfagaqbaaqabaGccqGH9aqpcaWGnb % GaamOtaiaac6cacaWGnbGaamyuaiaac6cacaWGnbGabmytayaafaaa % aa!45EE! {V_{MNPQ.M'N'P'Q'}} = MN.MQ.MM'\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Jaam % yqaiaadkeacaGGUaGaamyqaiaadseacaGGUaGaam4uaiaadIeacaGG % UaGaam4AamaaCaaaleqabaGaaGOmaaaakiaac6cadaqadaqaaiaaig % dacqGHsislcaWGRbaacaGLOaGaayzkaaaaaa!4487! = AB.AD.SH.{k^2}.\left( {1 - k} \right)\)

Mà \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGtbGaaiOlaiaadgeacaWGcbGaam4qaiaadseaaeqaaOGa % eyypa0ZaaSaaaeaacaaIXaaabaGaaG4maaaacaWGtbGaamisaiaac6 % cacaWGbbGaamOqaiaac6cacaWGbbGaamiraaaa!4460! {V_{S.ABCD}} = \frac{1}{3}SH.AB.AD\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OvamaaBaaaleaacaWGnbGaamOtaiaadcfacaWGrbGaaiOlaiqad2ea % gaqbaiqad6eagaqbaiqadcfagaqbaiqadgfagaqbaaqabaGccqGH9a % qpcaaIZaGaaiOlaiaadAfadaWgaaWcbaGaam4uaiaac6cacaWGbbGa % amOqaiaadoeacaWGebaabeaakiaac6cacaWGRbWaaWbaaSqabeaaca % aIYaaaaOGaaiOlamaabmaabaGaaGymaiabgkHiTiaadUgaaiaawIca % caGLPaaaaaa!507A! \Rightarrow {V_{MNPQ.M'N'P'Q'}} = 3.{V_{S.ABCD}}.{k^2}.\left( {1 - k} \right)\)

Thể tích khối chóp không đổi nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGnbGaamOtaiaadcfacaWGrbGaaiOlaiqad2eagaqbaiqa % d6eagaqbaiqadcfagaqbaiqadgfagaqbaaqabaaaaa!3E7D! {V_{MNPQ.M'N'P'Q'}}\) đạt giá trị lớn nhất khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4AamaaCa % aaleqabaGaaGOmaaaakiaac6cadaqadaqaaiaaigdacqGHsislcaWG % RbaacaGLOaGaayzkaaaaaa!3CAA! {k^2}.\left( {1 - k} \right)\) lớn nhất.

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4AamaaCa % aaleqabaGaaGOmaaaakiaac6cadaqadaqaaiaadUgacqGHsislcaaI % XaaacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaaIYaWaaeWaaeaaca % aIXaGaeyOeI0Iaam4AaaGaayjkaiaawMcaaiaac6cacaWGRbGaaiOl % aiaadUgaaeaacaaIYaaaaiabgsMiJoaalaaabaGaaGymaaqaaiaaik % daaaWaaeWaaeaadaWcaaqaaiaaikdacqGHsislcaaIYaGaam4Aaiab % gUcaRiaadUgacqGHRaWkcaWGRbaabaGaaG4maaaaaiaawIcacaGLPa % aadaahaaWcbeqaaiaaiodaaaGccqGH9aqpdaWcaaqaaiaaisdaaeaa % caaIYaGaaG4naaaaaaa!576C! {k^2}.\left( {k - 1} \right) = \frac{{2\left( {1 - k} \right).k.k}}{2} \le \frac{1}{2}{\left( {\frac{{2 - 2k + k + k}}{3}} \right)^3} = \frac{4}{{27}}\)

Đẳng thức xảy ra khi và chỉ khi: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmamaabm % aabaGaaGymaiabgkHiTiaadUgaaiaawIcacaGLPaaacqGH9aqpcaWG % Rbaaaa!3CC6! 2\left( {1 - k} \right) = k\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % 4Aaiabg2da9maalaaabaGaaGOmaaqaaiaaiodaaaaaaa!3BCE! \Leftrightarrow k = \frac{2}{3}\)Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGtbGaamytaaqaaiaadofacaWGbbaaaiabg2da9maalaaabaGaaGOm % aaqaaiaaiodaaaaaaa!3BDB! \frac{{SM}}{{SA}} = \frac{2}{3}\)

Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iqadAgagaqbamaabmaa % baGaamiEaaGaayjkaiaawMcaaiabgkHiTmaabmaabaGaaGOmaiaadI % hacqGHRaWkcaaIXaaacaGLOaGaayzkaaaaaa!43B6! h\left( x \right) = f'\left( x \right) - \left( {2x + 1} \right)\). Khi đó hàm số \(h(x)\) liên tục trên các đoạn \([-1;1] ;[1;2]\) ,  và có \(g(x)\) là một nguyên hàm của hàm số \(y = h(x)\)

Do đó diện tích hình phẳng giới hạn bởi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGH9aqpcqGHsislcaaIXaaabaGaamiEaiabg2da9iaa % igdaaeaacaWG5bGaeyypa0JabmOzayaafaWaaeWaaeaacaWG4baaca % GLOaGaayzkaaaabaGaamyEaiabg2da9iaaikdacaWG4bGaey4kaSIa % aGymaaaacaGL7baaaaa!485B! \left\{ \begin{array}{l} x = - 1\\ x = 1\\ y = f'\left( x \right)\\ y = 2x + 1 \end{array} \right.\) là

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaaIXaaabeaakiabg2da9maapehabaWaaqWaaeaaceWGMbGb % auaadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGHsisldaqadaqaai % aaikdacaWG4bGaey4kaSIaaGymaaGaayjkaiaawMcaaaGaay5bSlaa % wIa7aiaabsgacaWG4baaleaacqGHsislcaaIXaaabaGaaGymaaqdcq % GHRiI8aaaa!4BDE! {S_1} = \int\limits_{ - 1}^1 {\left| {f'\left( x \right) - \left( {2x + 1} \right)} \right|{\rm{d}}x} \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Zaa8 % qCaeaadaWadaqaaiqadAgagaqbamaabmaabaGaamiEaaGaayjkaiaa % wMcaaiabgkHiTmaabmaabaGaaGOmaiaadIhacqGHRaWkcaaIXaaaca % GLOaGaayzkaaaacaGLBbGaayzxaaGaaeizaiaadIhaaSqaaiabgkHi % TiaaigdaaeaacaaIXaaaniabgUIiYdaaaa!48E5! = \int\limits_{ - 1}^1 {\left[ {f'\left( x \right) - \left( {2x + 1} \right)} \right]{\rm{d}}x} \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Zaaq % GaaeaacaWGNbWaaeWaaeaacaWG4baacaGLOaGaayzkaaaacaGLiWoa % daqhaaWcbaGaeyOeI0IaaGymaaqaaiaaigdaaaaaaa!3E92! = \left. {g\left( x \right)} \right|_{ - 1}^1=g(1)-g(-1)\)

Vì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaaIXaaabeaakiabg6da+iaaicdaaaa!397F! {S_1} > 0\) nên \(g(1)>g(-1)\).

Diện tích hình phẳng giới hạn bởi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGH9aqpcaaIXaaabaGaamiEaiabg2da9iaaikdaaeaa % caWG5bGaeyypa0JabmOzayaafaWaaeWaaeaacaWG4baacaGLOaGaay % zkaaaabaGaamyEaiabg2da9iaaikdacaWG4bGaey4kaSIaaGymaaaa % caGL7baaaaa!476F! \left\{ \begin{array}{l} x = 1\\ x = 2\\ y = f'\left( x \right)\\ y = 2x + 1 \end{array} \right.\) là

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaaIYaaabeaakiabg2da9maapehabaWaaqWaaeaaceWGMbGb % auaadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGHsisldaqadaqaai % aaikdacaWG4bGaey4kaSIaaGymaaGaayjkaiaawMcaaaGaay5bSlaa % wIa7aiaabsgacaWG4baaleaacaaIXaaabaGaaGOmaaqdcqGHRiI8aa % aa!4AF3! {S_2} = \int\limits_1^2 {\left| {f'\left( x \right) - \left( {2x + 1} \right)} \right|{\rm{d}}x} \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Zaa8 % qCaeaadaWadaqaamaabmaabaGaaGOmaiaadIhacqGHRaWkcaaIXaaa % caGLOaGaayzkaaGaeyOeI0IabmOzayaafaWaaeWaaeaacaWG4baaca % GLOaGaayzkaaaacaGLBbGaayzxaaGaaeizaiaadIhaaSqaaiaaigda % aeaacaaIYaaaniabgUIiYdaaaa!47F9! = \int\limits_1^2 {\left[ {\left( {2x + 1} \right) - f'\left( x \right)} \right]{\rm{d}}x} \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Jaey % OeI0YaaqGaaeaacaWGNbWaaeWaaeaacaWG4baacaGLOaGaayzkaaaa % caGLiWoadaqhaaWcbaGaaGymaaqaaiaaikdaaaaaaa!3E93! = - \left. {g\left( x \right)} \right|_1^2=g(1) - g(2)\)

Vì \(S_{2}>0\) nên g(1) > g (2)

Gọi E là điểm đối xứng  của C qua điểm B . Khi đó tam giác ACE vuông tại A .

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yqaiaadweacqGH9aqpdaGcaaqaaiaaisdacaWGHbWaaWbaaSqabeaa % caaIYaaaaOGaeyOeI0IaamyyamaaCaaaleqabaGaaGOmaaaaaeqaaO % Gaeyypa0JaamyyamaakaaabaGaaG4maaWcbeaaaaa!4317! \Rightarrow AE = \sqrt {4{a^2} - {a^2}} = a\sqrt 3 \)

Mặt khác, ta có BC' = B'E = AB' nên tam giác AB'E vuông cân tại B'.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yqaiqadkeagaqbaiabg2da9maalaaabaGaamyqaiaadweaaeaadaGc % aaqaaiaaikdaaSqabaaaaaaa!3D66! \Rightarrow AB' = \frac{{AE}}{{\sqrt 2 }}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaWGHbWaaOaaaeaacaaIZaaaleqaaaGcbaWaaOaaaeaacaaI % Yaaaleqaaaaaaaa!39A8! = \frac{{a\sqrt 3 }}{{\sqrt 2 }}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaWGHbWaaOaaaeaacaaI2aaaleqaaaGcbaGaaGOmaaaaaaa!3990! = \frac{{a\sqrt 6 }}{2}\)

Suy ra: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiqadg % eagaqbaiabg2da9maakaaabaWaaeWaaeaadaWcaaqaaiaadggadaGc % aaqaaiaaiAdaaSqabaaakeaacaaIYaaaaaGaayjkaiaawMcaamaaCa % aaleqabaGaaGOmaaaakiabgkHiTiaadggadaahaaWcbeqaaiaaikda % aaaabeaakiabg2da9maalaaabaGaamyyamaakaaabaGaaGOmaaWcbe % aaaOqaaiaaikdaaaaaaa!4413! AA' = \sqrt {{{\left( {\frac{{a\sqrt 6 }}{2}} \right)}^2} - {a^2}} = \frac{{a\sqrt 2 }}{2}\)

Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9maalaaabaGaamyyamaakaaabaGaaGOmaaWcbeaaaOqaaiaaikda % aaGaaiOlamaalaaabaGaamyyamaaCaaaleqabaGaaGOmaaaakmaaka % aabaGaaG4maaWcbeaaaOqaaiaaisdaaaaaaa!3EA2! V = \frac{{a\sqrt 2 }}{2}.\frac{{{a^2}\sqrt 3 }}{4}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaWGHbWaaWbaaSqabeaacaaIZaaaaOWaaOaaaeaacaaI2aaa % leqaaaGcbaGaaGioaaaaaaa!3A8A! = \frac{{{a^3}\sqrt 6 }}{8}\)

Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadIhadaahaaWcbeqa % aiaaisdaaaGccqGHsislcaaI0aGaamiEamaaCaaaleqabaGaaG4maa % aakiabgUcaRiaaisdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4k % aSIaamyyaaaa!4552! g\left( x \right) = {x^4} - 4{x^3} + 4{x^2} + a\).

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4zayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaaGinaiaadIha % daahaaWcbeqaaiaaiodaaaGccqGHsislcaaIXaGaaGOmaiaadIhada % ahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI4aGaamiEaaaa!441C! g'\left( x \right) = 4{x^3} - 12{x^2} + 8x\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4zayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaaGimaaaa!3B32! ;g'\left( x \right) = 0\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaaG % inaiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIXaGaaGOm % aiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI4aGaamiEai % abg2da9iaaicdaaaa!43B4! \Leftrightarrow 4{x^3} - 12{x^2} + 8x = 0\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aam % qaaqaabeqaaiaadIhacqGH9aqpcaaIWaaabaGaamiEaiabg2da9iaa % igdaaeaacaWG4bGaeyypa0JaaGOmaaaacaGLBbaaaaa!418D! \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 1\\ x = 2 \end{array} \right.\)

Bảng biến thiên

Do \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiaad2 % gacqGHLjYScaWGnbGaeyOpa4JaaGimaaaa!3BFC! 2m \ge M > 0\) nên \(m > 0\) suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabgcMi5kaaicdacaaMc8UaaGPa % VlabgcGiIiaadIhacqGHiiIZdaWadaqaaiaaicdacaGG7aGaaGOmaa % Gaay5waiaaw2faaaaa!4675! g\left( x \right) \ne 0\,\,\forall x \in \left[ {0;2} \right]\)

Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamqaaqaabe % qaaiaadggacqGHRaWkcaaIXaGaeyipaWJaaGimaaqaaiaadggacqGH % +aGpcaaIWaaaaiaawUfaaiabgsDiBpaadeaaeaqabeaacaWGHbGaey % ipaWJaeyOeI0IaaGymaaqaaiaadggacqGH+aGpcaaIWaaaaiaawUfa % aaaa!4777! \left[ \begin{array}{l} a + 1 < 0\\ a > 0 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} a < - 1\\ a > 0 \end{array} \right.\)

Nếu \(a < -1\) thì \(M= -a\); \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabg2 % da9iabgkHiTiaadggacqGHsislcaaIXaaaaa!3B67! m = - a - 1\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacqGHshI3caaIYaWdaiaacIcapeGaeyOeI0IaamyyaiabgkHiTiaa % igdapaGaaiyka8qacqGHLjYScqGHsislcaWGHbGaeyi1HSTaamyyai % abgsMiJkabgkHiTiaaikdaaaa!4877! \Rightarrow 2( - a - 1) \ge - a \Leftrightarrow a \le - 2\)

Nếu \(a > 0\) thì \(M = a + 1\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabg2 % da9iaadggaaaa!38D2! ;m = a\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4naaa!3851! \Rightarrow \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiaadg % gacqGHLjYScaWGHbGaey4kaSIaaGymaaaa!3BDF! 2a \ge a + 1\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % yyaiabgwMiZkaaigdaaaa!3BB7! \Leftrightarrow a \ge 1\)

Do đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabgs % MiJkabgkHiTiaaikdaaaa!3A38! a \le - 2\) hoặc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabgw % MiZkaaigdaaaa!395B! a \ge 1\) , do a nguyên và thuộc đoạn \([-3;3]\) nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabgI % GiopaacmaabaGaeyOeI0IaaG4maiaacUdacqGHsislcaaIYaGaai4o % aiaaigdacaGG7aGaaGOmaiaacUdacaaIZaaacaGL7bGaayzFaaaaaa!4312! a \in \left\{ { - 3; - 2;1;2;3} \right\}\) .

Vậy có 5 giá trị của a thỏa mãn đề bài.

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGHbaacaGLxdcacqGH9aqpcqGHsisldaWhcaqaaiaadMgaaiaawEni % aiabgUcaRiaaikdadaWhcaqaaiaadQgaaiaawEniaiabgkHiTiaaio % dadaWhcaqaaiaadUgaaiaawEniaaaa!45B2! \overrightarrow a = - \overrightarrow i + 2\overrightarrow j - 3\overrightarrow k \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aa8 % HaaeaacaWGHbaacaGLxdcadaqadaqaaiabgkHiTiaaigdacaGG7aGa % aGOmaiaacUdacqGHsislcaaIZaaacaGLOaGaayzkaaaaaa!4200! \Rightarrow \overrightarrow a \left( { - 1;2; - 3} \right)\)

Ta có : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadk % eacqGH9aqpcaaIYaWaaOaaaeaacaaIYaaaleqaaaaa!3A1A! AB = 2\sqrt 2\)

Phương trình mặt cầu tâm C bán kính AB : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaabmaabaGaam % iEaiabgUcaRiaaigdacaaIWaaacaGLOaGaayzkaaWaaWbaaSqabeaa % caaIYaaaaOGaey4kaSYaaeWaaeaacaWG5bGaeyOeI0IaaGymaiaaiE % daaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkdaqa % daqaaiaadQhacqGHRaWkcaaI3aaacaGLOaGaayzkaaWaaWbaaSqabe % aacaaIYaaaaOGaeyypa0JaaGioaaaa!4A46! {\left( {x + 10} \right)^2} + {\left( {y - 17} \right)^2} + {\left( {z + 7} \right)^2} = 8\)

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaeyOeI0IaaGinaiaadIhadaahaaWcbeqaaiaaiodaaaGc % cqGHRaWkcaaI0aGaamiEaaaa!3E3D! y' = - 4{x^3} + 4x\) .

Cho \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGimaaaa!38BE! y' = 0\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaey % OeI0IaaGinaiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaaI % 0aGaamiEaiabg2da9iaaicdaaaa!4049! \Leftrightarrow - 4{x^3} + 4x = 0\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aam % qaaqaabeqaaiaadIhacqGH9aqpcaaIWaGaeyycI88aaeWaaeaacaaI % WaGaai4oaiaaiodaaiaawIcacaGLPaaaaeaacaWG4bGaeyypa0JaaG % ymaiabgIGiopaabmaabaGaaGimaiaacUdacaaIZaaacaGLOaGaayzk % aaaabaGaamiEaiabg2da9iabgkHiTiaaigdacqGHjiYZdaqadaqaai % aaicdacaGG7aGaaG4maaGaayjkaiaawMcaaaaacaGLBbaaaaa!5246! \Leftrightarrow \left[ \begin{array}{l} x = 0 \notin \left( {0;3} \right)\\ x = 1 \in \left( {0;3} \right)\\ x = - 1 \notin \left( {0;3} \right) \end{array} \right.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yEamaabmaabaGaaGimaaGaayjkaiaawMcaaiabg2da9iaaikdaaaa!3D54! \Rightarrow y\left( 0 \right) = 2 ; % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEamaabm % aabaGaaGymaaGaayjkaiaawMcaaiabg2da9iaaiodaaaa!3AF9! y\left( 1 \right) = 3; % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEamaabm % aabaGaaG4maaGaayjkaiaawMcaaiabg2da9iabgkHiTiaaiAdacaaI % Xaaaaa!3CA6! y\left( 3 \right) = - 61\)

Vậy giá trị lớn nhất của hàm số là 3.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyDamaaBa % aaleaacaaI4aaabeaakiabg2da9iaadwhadaWgaaWcbaGaaGymaaqa % baGccqGHRaWkcaaI3aGaamizaaaa!3D63! {u_8} = {u_1} + 7d\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaaG % OmaiaaiAdacaaMc8Uaeyypa0ZaaSaaaeaacaaIXaaabaGaaG4maaaa % cqGHRaWkcaaI3aGaamizaaaa!4071! \Leftrightarrow 26\, = \frac{1}{3} + 7d\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % izaiabg2da9maalaaabaGaaGymaiaaigdaaeaacaaIZaaaaaaa!3C82! \Leftrightarrow d = \frac{{11}}{3}\)

Gọi số phức \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEaiabg2 % da9iaadIhacqGHRaWkcaWGPbGaamyEaiaaykW7daqadaqaaiaadIha % caGGSaGaamyEaiabgIGiolabl2riHcGaayjkaiaawMcaaaaa!4477! z = x + iy\,\left( {x,y \in R } \right)\)

Ta có:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaada % qdaaqaaiaadQhaaaGaey4kaSIaaGOmaiabgkHiTiaadMgaaiaawEa7 % caGLiWoacqGH9aqpcaaI0aGaeyi1HS9aaqWaaeaadaqadaqaaiaadI % hacqGHRaWkcaaIYaaacaGLOaGaayzkaaGaey4kaSYaaeWaaeaacqGH % sislcaWG5bGaeyOeI0IaaGymaaGaayjkaiaawMcaaiaadMgaaiaawE % a7caGLiWoacqGH9aqpcaaI0aaaaa!51B5! \left| {\overline z + 2 - i} \right| = 4 \Leftrightarrow \left| {\left( {x + 2} \right) + \left( { - y - 1} \right)i} \right| = 4\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aae % WaaeaacaWG4bGaey4kaSIaaGOmaaGaayjkaiaawMcaamaaCaaaleqa % baGaaGOmaaaakiabgUcaRmaabmaabaGaamyEaiabgUcaRiaaigdaai % aawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGH9aqpcaaIXaGa % aGOnaaaa!45E1! \Leftrightarrow {\left( {x + 2} \right)^2} + {\left( {y + 1} \right)^2} = 16\)

Vậy tập hợp tất cả các điểm biểu diễn các số phức \(z\) thỏa mãn: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaada % qdaaqaaiaadQhaaaGaey4kaSIaaGOmaiabgkHiTiaadMgaaiaawEa7 % caGLiWoacqGH9aqpcaaI0aaaaa!3F63! \left| {\overline z + 2 - i} \right| = 4\) là đường tròn có tâm \(I ( -2; -1)\) và có bán kính \(R = 4\)

Ta có \(OA = |z|\) ; \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4taiaadk % eacqGH9aqpdaabdaqaamaabmaabaGaaGymaiabgUcaRiaadMgaaiaa % wIcacaGLPaaacaWG6baacaGLhWUaayjcSdGaeyypa0ZaaOaaaeaaca % aIYaaaleqaaOWaaqWaaeaacaWG6baacaGLhWUaayjcSdaaaa!46D2! OB = \left| {\left( {1 + i} \right)z} \right| = \sqrt 2 \left| z \right|\)\(; AB = \left| {\left( {1 + i} \right)z - z} \right| = \left| {iz} \right| = \left| z \right|\)

Suy ra \(\Delta AOB\) vuông cân tại A ( OA =OB và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4taiaadg % eadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWGbbGaamOqamaaCaaa % leqabaGaaGOmaaaakiabg2da9iaad+eacaWGcbWaaWbaaSqabeaaca % aIYaaaaaaa!3F6D! O{A^2} + A{B^2} = O{B^2}\) )

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacqqHuoarcaWGpbGaamyqaiaadkeaaeqaaOGaeyypa0ZaaSaa % aeaacaaIXaaabaGaaGOmaaaacaWGpbGaamyqaiaac6cacaWGbbGaam % Oqaiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaWaaqWaaeaacaWG % 6baacaGLhWUaayjcSdWaaWbaaSqabeaacaaIYaaaaOGaeyypa0JaaG % ioaaaa!4A98! {S_{\Delta OAB}} = \frac{1}{2}OA.AB = \frac{1}{2}{\left| z \right|^2} = 8\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aaq % WaaeaacaWG6baacaGLhWUaayjcSdGaeyypa0JaaGinaaaa!3E35! \Leftrightarrow \left| z \right| = 4\)

Gọi O, O' lần lượt là tâm của hai mặt đáy.Khi đó tứ giác COO'C' là hình bình hành và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4qayaafa % Gabm4tayaafaGaeyypa0ZaaSaaaeaacaWGbbGaam4qaaqaaiaaikda % aaGaeyypa0Jaamyyaaaa!3CF4! C'O' = \frac{{AC}}{2} = a\)

Do \(BD//B'D'\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OqaiaadseacaaMe8Uaae4laiaab+cacaaMe8+aaeWaaeaacaWGdbGa % bmOqayaafaGabmirayaafaaacaGLOaGaayzkaaaaaa!4257! \Rightarrow BD\;{\rm{//}}\;\left( {CB'D'} \right)\) nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamOqaiaadseacaGG7aGaam4qaiqadseagaqbaaGaayjkaiaa % wMcaaiabg2da9iaadsgadaqadaqaaiaad+eacaGG7aWaaeWaaeaaca % WGdbGabmOqayaafaGabmirayaafaaacaGLOaGaayzkaaaacaGLOaGa % ayzkaaGaeyypa0JaamizamaabmaabaGabm4qayaafaGaai4oamaabm % aabaGaam4qaiqadkeagaqbaiqadseagaqbaaGaayjkaiaawMcaaaGa % ayjkaiaawMcaaaaa!4E5A! d\left( {BD;CD'} \right) = d\left( {O;\left( {CB'D'} \right)} \right) = d\left( {C';\left( {CB'D'} \right)} \right)\) .

Ta có :\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiqadkeagaqbaiqadseagaqbaiabgwQiEjqadgeagaqbaiqadoea % gaqbaaqaaiqadkeagaqbaiqadseagaqbaiabgwQiEjaadoeaceWGdb % GbauaaaaGaay5EaaGaeyO0H4TabmOqayaafaGabmirayaafaGaeyyP % I41aaeWaaeaacaWGdbGaam4taiqad+eagaqbaiqadoeagaqbaaGaay % jkaiaawMcaaaaa!4B98! \left\{ \begin{array}{l} B'D' \bot A'C'\\ B'D' \bot CC' \end{array} \right. \Rightarrow B'D' \bot \left( {COO'C'} \right)\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aae % WaaeaacaWGdbGabmOqayaafaGabmirayaafaaacaGLOaGaayzkaaGa % eyyPI41aaeWaaeaacaWGdbGaam4taiqad+eagaqbaiqadoeagaqbaa % GaayjkaiaawMcaaaaa!42D4! \Rightarrow \left( {CB'D'} \right) \bot \left( {COO'C'} \right)$\)

Lại có : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGdbGabmOqayaafaGabmirayaafaaacaGLOaGaayzkaaGaeyykIC8a % aeWaaeaacaWGdbGaam4taiqad+eagaqbaiqadoeagaqbaaGaayjkai % aawMcaaiabg2da9iaadoeaceWGpbGbauaaaaa!4312! \left( {CB'D'} \right) \cap \left( {COO'C'} \right) = CO'\)

Trong \(\Delta CC'O'\) hạ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4qayaafa % GaamisaiabgwQiEjaadoeaceWGpbGbauaacqGHshI3ceWGdbGbauaa % caWGibGaeyyPI41aaeWaaeaacaWGdbGabmOqayaafaGabmirayaafa % aacaGLOaGaayzkaaaaaa!4496! C'H \bot CO' \Rightarrow C'H \bot \left( {CB'D'} \right)\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % izamaabmaabaGaamOqaiaadseacaGG7aGaaGPaVlaadoeaceWGebGb % auaaaiaawIcacaGLPaaacqGH9aqpceWGdbGbauaacaWGibaaaa!42E1! \Rightarrow d\left( {BD;\,CD'} \right) = C'H\)

Khi đó : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGabm4qayaafaGaamisamaaCaaaleqabaGaaGOmaaaaaaGc % cqGH9aqpdaWcaaqaaiaaigdaaeaacaWGdbGabm4qayaafaWaaWbaaS % qabeaacaaIYaaaaaaakiabgUcaRmaalaaabaGaaGymaaqaaiqadoea % gaqbaiqad+eagaqbamaaCaaaleqabaGaaGOmaaaaaaGccqGH9aqpda % WcaaqaaiaaigdaaeaadaqadaqaaiaaikdacaWGHbaacaGLOaGaayzk % aaWaaWbaaSqabeaacaaIYaaaaaaakiabgUcaRmaalaaabaGaaGymaa % qaaiaadggadaahaaWcbeqaaiaaikdaaaaaaOGaeyypa0ZaaSaaaeaa % caaI1aaabaGaaGinaiaadggadaahaaWcbeqaaiaaikdaaaaaaaaa!4FDE! \frac{1}{{C'{H^2}}} = \frac{1}{{C{{C'}^2}}} + \frac{1}{{C'{{O'}^2}}} = \frac{1}{{{{\left( {2a} \right)}^2}}} + \frac{1}{{{a^2}}} = \frac{5}{{4{a^2}}}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Tabm % 4qayaafaGaamisaiabg2da9maalaaabaGaaGOmamaakaaabaGaaGyn % aaWcbeaakiaadggaaeaacaaI1aaaaaaa!3E4D! \Rightarrow C'H = \frac{{2\sqrt 5 a}}{5}\)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGHRaWkcaaI % Xaaaaa!3D00! t = f\left( x \right) + 1\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % iDaiabg2da9iaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaI% ZaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiAdacaWG4b % Gaey4kaSIaaGOmaaaa!4422! \Rightarrow t = {x^3} - 3{x^2} - 6x + 2\)

Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaOaaaeaaca % WGMbWaaeWaaeaacaWGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaaGa % ey4kaSIaaGymaaGaayjkaiaawMcaaiabgUcaRiaaigdaaSqabaGccq % GH9aqpcaWGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaaGaey4kaSIa % aGOmaaaa!454C! \sqrt {f\left( {f\left( x \right) + 1} \right) + 1} = f\left( x \right) + 2\) trở thành:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaOaaaeaaca % WGMbWaaeWaaeaacaWG0baacaGLOaGaayzkaaGaey4kaSIaaGymaaWc % beaakiabg2da9iaadshacqGHRaWkcaaIXaaaaa!3EBE! \sqrt {f\left( t \right) + 1} = t + 1\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaadshacqGHLjYScqGHsislcaaIXaaabaGaamOzamaa % bmaabaGaamiDaaGaayjkaiaawMcaaiabgUcaRiaaigdacqGH9aqpca % WG0bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiaadshacqGH % RaWkcaaIXaaaaiaawUhaaaaa!4A07! \Leftrightarrow \left\{ \begin{array}{l} t \ge - 1\\ f\left( t \right) + 1 = {t^2} + 2t + 1 \end{array} \right.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaadshacqGHLjYScqGHsislcaaIXaaabaGaamiDamaa % CaaaleqabaGaaG4maaaakiabgkHiTiaaisdacaWG0bWaaWbaaSqabe % aacaaIYaaaaOGaeyOeI0IaaGioaiaadshacqGHRaWkcaaIXaGaeyyp % a0JaaGimaaaacaGL7baaaaa!4960! \Leftrightarrow \left\{ \begin{array}{l} t \ge - 1\\ {t^3} - 4{t^2} - 8t + 1 = 0 \end{array} \right.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaadshacqGHLjYScqGHsislcaaIXaaabaWaamqaaqaa % beqaaiaadshacqGH9aqpcaWG0bWaaSbaaSqaaiaaigdaaeqaaOGaey % icI48aaeWaaeaacqGHsislcaaIYaGaai4oaiabgkHiTiaaigdaaiaa % wIcacaGLPaaaaeaacaWG0bGaeyypa0JaamiDamaaBaaaleaacaaIYa % aabeaakiabgIGiopaabmaabaGaeyOeI0IaaGymaiaacUdacaaMc8Ua % aGymaaGaayjkaiaawMcaaaqaaiaadshacqGH9aqpcaWG0bWaaSbaaS % qaaiaaiodaaeqaaOGaeyicI48aaeWaaeaacaaIXaGaai4oaiaaykW7 % caaI2aaacaGLOaGaayzkaaaaaiaawUfaaaaacaGL7baaaaa!6041! \Leftrightarrow \left\{ \begin{array}{l} t \ge - 1\\ \left[ \begin{array}{l} t = {t_1} \in \left( { - 2; - 1} \right)\\ t = {t_2} \in \left( { - 1;\,1} \right)\\ t = {t_3} \in \left( {1;\,6} \right) \end{array} \right. \end{array} \right.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aam % qaaqaabeqaaiaadshacqGH9aqpcaWG0bWaaSbaaSqaaiaaikdaaeqa % aOGaeyicI48aaeWaaeaacqGHsislcaaIXaGaai4oaiaaykW7caaIXa % aacaGLOaGaayzkaaaabaGaamiDaiabg2da9iaadshadaWgaaWcbaGa % aG4maaqabaGccqGHiiIZdaqadaqaaiaaiwdacaGG7aGaaGPaVlaaiA % daaiaawIcacaGLPaaaaaGaay5waaaaaa!4FB6! \Leftrightarrow \left[ \begin{array}{l} t = {t_2} \in \left( { - 1;\,1} \right)\\ t = {t_3} \in \left( {5;\,6} \right) \end{array} \right.\)

Vì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiDaaGaayjkaiaawMcaaiabg2da9iaadshadaahaaWcbeqa % aiaaiodaaaGccqGHsislcaaI0aGaamiDamaaCaaaleqabaGaaGOmaa % aakiabgkHiTiaaiIdacaWG0bGaey4kaSIaaGymaaaa!4430! g\left( t \right) = {t^3} - 4{t^2} - 8t + 1\); \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaeyOeI0IaaGOmaaGaayjkaiaawMcaaiabg2da9iabgkHiTiaa % iEdaaaa!3CC5! g\left( { - 2} \right) = - 7\); \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaeyOeI0IaaGymaaGaayjkaiaawMcaaiabg2da9iaaisdaaaa!3BD4! g\left( { - 1} \right) = 4\); \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaaGymaaGaayjkaiaawMcaaiabg2da9iabgkHiTiaaigdacaaI % Waaaaa!3C8B! g\left( 1 \right) = - 10\); \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaaGOnaaGaayjkaiaawMcaaiabg2da9iaaikdacaaI1aaaaa!3BA9! g\left( 6 \right) = 25\) ; \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaaGynaaGaayjkaiaawMcaaiabg2da9iabgkHiTiaaigdacaaI % 0aaaaa!3C93! g\left( 5 \right) = - 14\); .

Xét \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIZaGaamiE % amaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiAdacaWG4bGaey4kaS % IaaGOmaaaa!41C5! t = {x^3} - 3{x^2} - 6x + 2\)

 Ta có

Dựa vào bảng biến thiên, ta có

+ Với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaadshadaWgaaWcbaGaaGOmaaqabaGccqGHiiIZdaqadaqaaiab % gkHiTiaaigdacaGG7aGaaGPaVlaaigdaaiaawIcacaGLPaaaaaa!4197! t = {t_2} \in \left( { - 1;\,1} \right)\), ta có d cắt tại 3 điểm phân biệt, nên phương trình có 3 nghiệm.

+ Với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaadshadaWgaaWcbaGaaG4maaqabaGccqGHiiIZdaqadaqaaiaa % iwdacaGG7aGaaGPaVlaaiAdaaiaawIcacaGLPaaaaaa!40B4! t = {t_3} \in \left( {5;\,6} \right)\), ta có d cắt tại 1 điểm, nên phương trình có 1 nghiệm.

Vậy phương trình đã cho có 4 nghiệm

Thể tích khối trụ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9iabec8aWjaadkhadaahaaWcbeqaaiaaikdaaaGccaWGObGaeyyp % a0JaeqiWdaNaaiOlaiaaikdadaahaaWcbeqaaiaaikdaaaGccaGGUa % GaaGOmaiabg2da9iaaiIdacqaHapaCaaa!467F! V = \pi {r^2}h = \pi {.2^2}.2 = 8\pi \)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaW2dcaWG0b % Gaeyypa0JaaGOmamaaCaaaleqabaGaamiEaaaaaaa!3A56! t= {2^x}; t > 0\), . Phương trình trở thành: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaW2dcaWG0b % WaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOmaiaad2gacaWG0bGa % ey4kaSIaaGOmaiaad2gacqGH9aqpcaaIWaaaaa!4041! {t^2} - 2mt + 2m = 0\)  (1)

Phương trình đã cho có hai nghiệm \(x_{1} ; x_{2}\) ,  thỏa mãn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaW2dcaWG4b % WaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaamiEamaaBaaaleaacaaI % Yaaabeaakiabg2da9iaaiodaaaa!3CF3! {x_1} + {x_2} = 3\) khi và chỉ khi phương trình (1) có hai nghiệm dương phân biệt thỏa mãn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaW2dcaWG0b % WaaSbaaSqaaiaaigdaaeqaaOGaaiOlaiaadshadaWgaaWcbaGaaGOm % aaqabaGccqGH9aqpcaaIYaWaaWbaaSqabeaacaWG4bWaaSbaaWqaai % aaigdaaeqaaaaakiaac6cacaaIYaWaaWbaaSqabeaacaWG4bWaaSba % aWqaaiaaikdaaeqaaaaakiabg2da9iaaikdadaahaaWcbeqaaiaadI % hadaWgaaadbaGaaGymaaqabaWccqGHRaWkcaWG4bWaaSbaaWqaaiaa % ikdaaeqaaaaakiabg2da9iaaikdadaahaaWcbeqaaiaaiodaaaGccq % GH9aqpcaaI4aaaaa!4D90! {t_1}.{t_2} = {2^{{x_1}}}{.2^{{x_2}}} = {2^{{x_1} + {x_2}}} = {2^3} = 8\)

Khi đó phương trình (1) có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaW2ddaGaba % abceqabaaBpeaacuqHuoargaqbaiabg2da9iaad2gadaahaaWcbeqa % aiaaikdaaaGccqGHsislcaaIYaGaamyBaiabg6da+iaaicdaaeaaca % WGtbGaeyypa0JaaGOmaiaad2gacqGH+aGpcaaIWaaabaGaamiuaiab % g2da9iaaikdacaWGTbGaeyOpa4JaaGimaaqaaiaadcfacqGH9aqpca % aIYaGaamyBaiabg2da9iaaiIdaaaGaay5EaaGaeyi1HSTaamyBaiab % g2da9iaaisdaaaa!55C8! \left\{ \begin{array}{l} \Delta ' = {m^2} - 2m > 0\\ S = 2m > 0\\ P = 2m > 0\\ P = 2m = 8 \end{array} \right. \Leftrightarrow m = 4\)

Số phần tử của không gian mẫu là số cách chọn 4 đỉnh trong 32 đỉnh để tạo thành tứ giác, \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaacq % GHPoWvaiaawEa7caGLiWoacqGH9aqpcaWGdbWaa0baaSqaaiaaioda % caaIYaaabaGaaGinaaaaaaa!3ED7! \left| \Omega \right| = C_{32}^4\)

Gọi A là biến cố "chọn được hình chữ nhật".

Để chọn được hình chữ nhật cần chọn 4 trong 16 đường chéo đi qua tâm của đa giác, do đó số phần tử của A là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qamaaDa % aaleaacaaIXaGaaGOnaaqaaiaaikdaaaaaaa!3920! C_{16}^2\).

Xác suất biến cố A là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaabm % aabaGaamyqaaGaayjkaiaawMcaaiabg2da9maalaaabaGaam4qamaa % DaaaleaacaaIXaGaaGOnaaqaaiaaikdaaaaakeaacaWGdbWaa0baaS % qaaiaaiodacaaIYaaabaGaaGinaaaaaaaaaa!4090! P\left( A \right) = \frac{{C_{16}^2}}{{C_{32}^4}}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaaIZaaabaGaaGioaiaaiMdacaaI5aaaaaaa!3A0F! = \frac{3}{{899}}\)

Tập xác định \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVC0df9qqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiraiabg2 % da9iabl2riHkaacYfadaGadaqaaiabgkHiTiaad2gaaiaawUhacaGL % 9baaaaa!3D85! D = R\backslash \left\{ { - m} \right\}\) D = R \ {-m}. Ta có \(y' = \frac{{{m^2} - 4}}{{{{\left( {x + m} \right)}^2}}}\) . Hàm số nghịch biến trên khoảng \(\left( { - \infty ;1} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTabm % yEayaafaGaeyipaWJaaGimaaaa!3B17! \Leftrightarrow y' < 0\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiaIiIaam % iEaiabgIGiopaabmaabaGaeyOeI0IaeyOhIuQaai4oaiaaigdaaiaa % wIcacaGLPaaacqGHuhY2daGabaabaeqabaGaamyBamaaCaaaleqaba % GaaGOmaaaakiabgkHiTiaaisdacqGH8aapcaaIWaaabaGaaGymaiab % gsMiJkabgkHiTiaad2gaaaGaay5Eaaaaaa!4BBF! \forall x \in \left( { - \infty ;1} \right) \Leftrightarrow \left\{ \begin{array}{l} {m^2} - 4 < 0\\ 1 \le - m \end{array} \right.\)\(y'<0, \forall x \in (-\infty;1)\)\(\Rightarrow m^2 - 4 <0 \iff - 2 < m < 2\)

Mặt khác \(\begin{cases} \forall x \in (-\infty;1) \\ x + m \neq 0 \end{cases}\)\(\iff \begin{cases} -x \in ( -1 ; +\infty) \\ m\neq -x \end{cases}\)

\(\Rightarrow m \in(-\infty;-1]\Rightarrow -2 < m \le -1\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0ZaaSaaaeaacaWGLbWaaWbaaSqabeaacaWG4baaaaGcbaGa % amyzamaaCaaaleqabaGaamiEaaaakiabgUcaRiaad2gadaahaaWcbe % qaaiaaikdaaaaaaOGaeyO0H4TabmyEayaafaWaaeWaaeaacaaIXaaa % caGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaWGLbaabaGaamyzaiabgU % caRiaad2gadaahaaWcbeqaaiaaikdaaaaaaaaa!4A69! y' = \frac{{{e^x}}}{{{e^x} + {m^2}}} \Rightarrow y'\left( 1 \right) = \frac{e}{{e + {m^2}}}\).

Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % WaaeWaaeaacaaIXaaacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaaI % XaaabaGaaGOmaaaacqGHuhY2daWcaaqaaiaadwgaaeaacaWGLbGaey % 4kaSIaamyBamaaCaaaleqabaGaaGOmaaaaaaGccqGH9aqpdaWcaaqa % aiaaigdaaeaacaaIYaaaaiabgsDiBlaaikdacaWGLbGaeyypa0Jaam % yzaiabgUcaRiaad2gadaahaaWcbeqaaiaaikdaaaGccqGHuhY2caWG % TbGaeyypa0JaeyySae7aaOaaaeaacaWGLbaaleqaaaaa!5563! y'\left( 1 \right) = \frac{1}{2} \Leftrightarrow \frac{e}{{e + {m^2}}} = \frac{1}{2} \Leftrightarrow 2e = e + {m^2} \Leftrightarrow m = \pm \sqrt e \).

Cách 1: Sử dụng tích phân từng phần ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGacaGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9maapeaabaGaamiEaiaadwgadaahaaWcbeqaaiaadIhaaaaabeqa % b0Gaey4kIipakiaabsgacaWG4bGaeyypa0Zaa8qaaeaacaWG4baale % qabeqdcqGHRiI8aOGaaGPaVlaabsgacaWGLbWaaWbaaSqabeaacaWG % 4baaaOGaeyypa0JaamiEaiaadwgadaahaaWcbeqaaiaadIhaaaGccq % GHsisldaWdbaqaaiaadwgadaahaaWcbeqaaiaadIhaaaaabeqab0Ga % ey4kIipakiaabsgacaWG4bGaeyypa0JaamiEaiaadwgadaahaaWcbe % qaaiaadIhaaaGccqGHsislcaWGLbWaaWbaaSqabeaacaWG4baaaOGa % ey4kaSIaam4qaiaac6caaaa!5BD4! I = \int {x{e^x}} {\rm{d}}x = \int x \,{\rm{d}}{e^x} = x{e^x} - \int {{e^x}} {\rm{d}}x = x{e^x} - {e^x} + C.\)

Cách 2: Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGacaGaaiaabeqaamaabaabaaGcbaGabmysayaafa % Gaeyypa0ZaaeWaaeaacaWG4bGaamyzamaaCaaaleqabaGaamiEaaaa % kiabgkHiTiaadwgadaahaaWcbeqaaiaadIhaaaGccqGHRaWkcaWGdb % aacaGLOaGaayzkaaWaaWbaaSqabeaakiadacUHYaIOaaGaeyypa0Ja % amyzamaaCaaaleqabaGaamiEaaaakiabgUcaRiaadIhacaWGLbWaaW % baaSqabeaacaWG4baaaOGaeyOeI0IaamyzamaaCaaaleqabaGaamiE % aaaakiabg2da9iaadIhacaWGLbWaaWbaaSqabeaacaWG4baaaOGaai % Olaaaa!5348! I' = {\left( {x{e^x} - {e^x} + C} \right)^\prime } = {e^x} + x{e^x} - {e^x} = x{e^x}.\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaaGimaiabgsDi % BpaadeaaeaqabeaacaWG4bGaeyypa0JaeyOeI0IaaGymaaqaaiaadI % hacqGH9aqpcaaIYaaabaGaamiEaiabg2da9iabgkHiTiaaiodaaaGa % ay5waaaaaa!48A7! f'\left( x \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = - 1\\ x = 2\\ x = - 3 \end{array} \right.\).

Ta có bảng biến thiên của hàm số \(f(x)\) :

Ta có bảng biến thiên của hàm số \(f(|x|)\):

Dựa vào bảng biến thiên ta thấy số điểm cực trị của hàm số \(f(|x|)\) là 3 .

Giả sử z = a + bi ; w = x + yi ; \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGHbGaaiilaiaadkgacaGGSaGaamiEaiaacYcacaWG5bGaeyicI4Sa % eSyhHekacaGLOaGaayzkaaaaaa!4049! \left( {a,b,x,y \in R} \right)\)  . Ta có 

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG6bGaeyOeI0IaaG4maiabgkHiTiaaikdacaWGPbaacaGLhWUaayjc % SdGaeyizImQaaGymaaaa!40C6! \left| {z - 3 - 2i} \right| \le 1\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aae % WaaeaacaWGHbGaeyOeI0IaaG4maaGaayjkaiaawMcaamaaCaaaleqa % baGaaGOmaaaakiabgUcaRmaabmaabaGaamOyaiabgkHiTiaaikdaai % aawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHKjYOcaaIXaaa % aa!45BA! \Leftrightarrow {\left( {a - 3} \right)^2} + {\left( {b - 2} \right)^2} \le 1\)

. Suy ra tập hợp điểm M biểu diễn số phức z là hình tròn tâm I ( 3;2), bán kính R = 1.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG3bGaey4kaSIaaGymaiabgUcaRiaaikdacaWGPbaacaGLhWUaayjc % SdGaeyizIm6aaqWaaeaacaWG3bGaeyOeI0IaaGOmaiabgkHiTiaadM % gaaiaawEa7caGLiWoaaaa!4792! \left| {w + 1 + 2i} \right| \le \left| {w - 2 - i} \right|\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aae % WaaeaacaWG4bGaey4kaSIaaGymaaGaayjkaiaawMcaamaaCaaaleqa % baGaaGOmaaaakiabgUcaRmaabmaabaGaamyEaiabgUcaRiaaikdaai % aawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHKjYOdaqadaqa % aiaadIhacqGHsislcaaIYaaacaGLOaGaayzkaaWaaWbaaSqabeaaca % aIYaaaaOGaey4kaSYaaeWaaeaacaWG5bGaeyOeI0IaaGymaaGaayjk % aiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgsDiBlaadIhacqGHR % WkcaWG5bGaeyizImQaaGimaaaa!57E3! \Leftrightarrow {\left( {x + 1} \right)^2} + {\left( {y + 2} \right)^2} \le {\left( {x - 2} \right)^2} + {\left( {y - 1} \right)^2} \Leftrightarrow x + y \le 0\)

 Suy ra tập hợp điểm N biểu diễn số phức w là nửa mặt phẳng giới hạn bởi đường thẳng \(\Delta: x + y = 0\) không chứa I

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamysaiaacYcacqqHuoaraiaawIcacaGLPaaacqGH9aqpdaWc % aaqaaiaaiwdaaeaadaGcaaqaaiaaikdaaSqabaaaaaaa!3DF6! d\left( {I,\Delta } \right) = \frac{5}{{\sqrt 2 }}\). Gọi H là hình chiếu của I trên \(\Delta\) .

Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG6bGaeyOeI0Iaam4DaaGaay5bSlaawIa7aiabg2da9iaad2eacaWG % obGaeyyzImRaamizamaabmaabaGaamysaiaacYcacqqHuoaraiaawI % cacaGLPaaacqGHsislcaWGsbGaeyypa0ZaaSaaaeaacaaI1aWaaOaa % aeaacaaIYaaaleqaaaGcbaGaaGOmaaaacqGHsislcaaIXaaaaa!4CA3! \left| {z - w} \right| = MN \ge d\left( {I,\Delta } \right) - R = \frac{{5\sqrt 2 }}{2} - 1\). Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaaBa % aaleaaciGGTbGaaiyAaiaac6gaaeqaaOGaeyypa0ZaaSaaaeaacaaI % 1aWaaOaaaeaacaaIYaaaleqaaaGcbaGaaGOmaaaacqGHsislcaaIXa % aaaa!3EEB! {P_{\min }} = \frac{{5\sqrt 2 }}{2} - 1\).

Hàm số xác định khi: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgk % HiTiaaigdacqGH+aGpcaaIWaGaeyi1HSTaamiEaiabg6da+iaaigda % aaa!3F77! x - 1 > 0 \Leftrightarrow x > 1\). Vậy tập xác định: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiraiabg2 % da9maabmaabaGaaGymaiaacUdacaaMc8Uaey4kaSIaeyOhIukacaGL % OaGaayzkaaaaaa!3EA4! D = \left( {1;\, + \infty } \right)\).

Nguyên hàm không có tính chất nguyên hàm của tích bằng tích các nguyên hàm.

Hoặc A, C, D đúng do đó là các tính chất cơ bản của nguyên hàm nên B sai

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaaikdacaWG5b % WaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaaG4naiaadMhacqGHRaWk % caaIYaGaamiEamaakaaabaGaaGymaiabgkHiTiaadIhaaSqabaGccq % GH9aqpcaaIZaWaaOaaaeaacaaIXaGaeyOeI0IaamiEaaWcbeaakiab % gUcaRiaaiodadaqadaqaaiaaikdacaWG5bWaaWbaaSqabeaacaaIYa % aaaOGaey4kaSIaaGymaaGaayjkaiaawMcaaaaa!4C9C! 2{y^3} + 7y + 2x\sqrt {1 - x} = 3\sqrt {1 - x} + 3\left( {2{y^2} + 1} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgsDiBlaaik % dadaqadaqaaiaadMhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaI % ZaGaamyEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiodacaWG5b % GaeyOeI0IaaGymaaGaayjkaiaawMcaaiabgUcaRmaabmaabaGaamyE % aiabgkHiTiaaigdaaiaawIcacaGLPaaacqGH9aqpcaaIYaWaaeWaae % aacaaIXaGaeyOeI0IaamiEaaGaayjkaiaawMcaamaakaaabaGaaGym % aiabgkHiTiaadIhaaSqabaGccqGHRaWkcaaIZaWaaOaaaeaacaaIXa % GaeyOeI0IaamiEaaWcbeaakiabgkHiTiaaikdadaGcaaqaaiaaigda % cqGHsislcaWG4baaleqaaaaa!5AF9! \Leftrightarrow 2\left( {{y^3} - 3{y^2} + 3y - 1} \right) + \left( {y - 1} \right) = 2\left( {1 - x} \right)\sqrt {1 - x} + 3\sqrt {1 - x} - 2\sqrt {1 - x} \)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgsDiBlaaik % dadaqadaqaaiaadMhacqGHsislcaaIXaaacaGLOaGaayzkaaWaaWba % aSqabeaacaaIZaaaaOGaey4kaSYaaeWaaeaacaWG5bGaeyOeI0IaaG % ymaaGaayjkaiaawMcaaiabg2da9iaaikdadaqadaqaamaakaaabaGa % aGymaiabgkHiTiaadIhaaSqabaaakiaawIcacaGLPaaadaahaaWcbe % qaaiaaiodaaaGccqGHRaWkdaGcaaqaaiaaigdacqGHsislcaWG4baa % leqaaOGaaGPaVlaaykW7daqadaqaaiaaigdaaiaawIcacaGLPaaaaa % a!5344! \Leftrightarrow 2{\left( {y - 1} \right)^3} + \left( {y - 1} \right) = 2{\left( {\sqrt {1 - x} } \right)^3} + \sqrt {1 - x} \,\,\left( 1 \right)\)

Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadAgadaqada % qaaiaadshaaiaawIcacaGLPaaacqGH9aqpcaaIYaGaamiDamaaCaaa % leqabaGaaG4maaaakiabgUcaRiaadshaaaa!3EE0! f\left( t \right) = 2{t^3} + t\) trên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaajibabaGaaG % imaiaacUdacaaMc8Uaey4kaSIaeyOhIukacaGLBbGaayzkaaaaaa!3D13! \left[ {0;\, + \infty } \right)\).

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiqadAgagaqbam % aabmaabaGaamiDaaGaayjkaiaawMcaaiabg2da9iaaiAdacaWG0bWa % aWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGymaaaa!3EB1! f'\left( t \right) = 6{t^2} + 1 > 0\) với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgcGiIiaads % hacqGHLjYScaaIWaaaaa!3A32! \forall t \ge 0\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkDiElaadA % gadaqadaqaaiaadshaaiaawIcacaGLPaaaaaa!3BB3! \Rightarrow f\left( t \right)\) luôn đồng biến trên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaajibabaGaaG % imaiaacUdacaaMc8Uaey4kaSIaeyOhIukacaGLBbGaayzkaaaaaa!3D13! \left[ {0;\, + \infty } \right)\).

Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaabmaabaGaaG % ymaaGaayjkaiaawMcaaiabgsDiBlaadMhacqGHsislcaaIXaGaeyyp % a0ZaaOaaaeaacaaIXaGaeyOeI0IaamiEaaWcbeaaaaa!40F5! \left( 1 \right) \Leftrightarrow y - 1 = \sqrt {1 - x} \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgsDiBlaadM % hacqGH9aqpcaaIXaGaey4kaSYaaOaaaeaacaaIXaGaeyOeI0IaamiE % aaWcbeaaaaa!3EA6! \Leftrightarrow y = 1 + \sqrt {1 - x} \).

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabgkDiElaadc % facqGH9aqpcaWG4bGaey4kaSIaaGOmaiaadMhacqGH9aqpcaWG4bGa % ey4kaSIaaGOmaiabgUcaRiaaikdadaGcaaqaaiaaigdacqGHsislca % WG4baaleqaaaaa!45B9! \Rightarrow P = x + 2y = x + 2 + 2\sqrt {1 - x} \) với  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaabmaabaGaam % iEaiabgsMiJkaaigdaaiaawIcacaGLPaaaaaa!3ADF! \left( {x \le 1} \right)\).

Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiaadEgadaqada % qaaiaadIhaaiaawIcacaGLPaaacqGH9aqpcaaIYaGaey4kaSIaamiE % aiabgUcaRiaaikdadaGcaaqaaiaaigdacqGHsislcaWG4baaleqaaa % aa!415A! g\left( x \right) = 2 + x + 2\sqrt {1 - x} \) trên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaajadabaGaey % OeI0IaeyOhIuQaai4oaiaaykW7caaIXaaacaGLOaGaayzxaaaaaa!3D3E! \left( { - \infty ;\,1} \right]\).

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiqadEgagaqbam % aabmaabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaigdacqGHsisl % daWcaaqaaiaaigdaaeaadaGcaaqaaiaaigdacqGHsislcaWG4baale % qaaaaaaaa!3FA0! g'\left( x \right) = 1 - \frac{1}{{\sqrt {1 - x} }}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiabg2da9maala % aabaWaaOaaaeaacaaIXaGaeyOeI0IaamiEaaWcbeaakiabgkHiTiaa % igdaaeaadaGcaaqaaiaaigdacqGHsislcaWG4baaleqaaaaaaaa!3E31! = \frac{{\sqrt {1 - x} - 1}}{{\sqrt {1 - x} }}\) .\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaaiqadEgagaqbam % aabmaabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaicdacqGHshI3 % caWG4bGaeyypa0JaaGimaaaa!4041! g'\left( x \right) = 0 \Rightarrow x = 0\)

Bảng biến thiên g(x):

Từ bảng biến thiên của hàm số g(x) suy ra giá trị lớn nhất của P là: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVCI8FfYJH8YrFfeuY-Hhbbf9v8qqaqFr0xc9pk0xbb % a9q8WqFfeaY-biLkVcLq-JHqpepeea0-as0Fb9pgeaYRXxe9vr0-vr % 0-vqpWqaaeaabiGaciaacaqabeaadaqaaqaaaOqaamaaxababaGaci % yBaiaacggacaGG4baaleaadaqcWaqaaiabgkHiTiabg6HiLkaacUda % caaMc8UaaGymaaGaayjkaiaaw2faaaqabaGccaWGNbWaaeWaaeaaca % WG4baacaGLOaGaayzkaaGaeyypa0JaaGinaaaa!458B! \mathop {\max }\limits_{\left( { - \infty ;\,1} \right]} g\left( x \right) = 4\)

Vì hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaamiEaiabgkHiTiaaikdaaeaacaWG4bGaeyOeI0Ia % aGymaaaaaaa!3D53! y = \frac{{x - 2}}{{x - 1}}\) có tập xác định \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiraiabg2 % da9iabl2riHkaacYfadaGadaqaaiaaigdaaiaawUhacaGL9baaaaa!3CFF! D = R\backslash \left\{ 1 \right\}\) nên hàm số không đồng biến trên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaacq % GHsislcqGHEisPcaGG7aGaey4kaSIaeyOhIukacaGLOaGaayzkaaaa % aa!3CED! \left( { - \infty ; + \infty } \right)\)

Dựa vào bảng biến thiên ta thấy phương trình có 4 nghiệm phân biệt khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % 4maiabgYda8iaad2gacqGH8aapcaaIYaaaaa!3B54! - 3 < m < 2\).

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGinaiaadQ % hadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaIXaGaaGOnaiaadQha % cqGHRaWkcaaIXaGaaG4naiabg2da9iaaicdacqGHuhY2daWabaabae % qabaGaamOEamaaBaaaleaacaaIXaaabeaakiabg2da9iaaikdacqGH % sisldaWcaaqaaiaaigdaaeaacaaIYaaaaiaadMgaaeaacaWG6bWaaS % baaSqaaiaaikdaaeqaaOGaeyypa0JaaGOmaiabgUcaRmaalaaabaGa % aGymaaqaaiaaikdaaaGaamyAaaaacaGLBbaaaaa!51A5! 4{z^2} - 16z + 17 = 0 \Leftrightarrow \left[ \begin{array}{l} {z_1} = 2 - \frac{1}{2}i\\ {z_2} = 2 + \frac{1}{2}i \end{array} \right.\)

Khi đó: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4Daiabg2 % da9maabmaabaGaaGymaiabgUcaRiaaikdacaWGPbaacaGLOaGaayzk % aaGaamOEamaaBaaaleaacaaIXaaabeaakiabgkHiTmaalaaabaGaaG % 4maaqaaiaaikdaaaGaamyAaaaa!4219! w = \left( {1 + 2i} \right){z_1} - \frac{3}{2}i\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Zaae % WaaeaacaaIXaGaey4kaSIaaGOmaiaadMgaaiaawIcacaGLPaaadaqa % daqaaiaaikdacqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaaaiaadM % gaaiaawIcacaGLPaaacqGHsisldaWcaaqaaiaaiodaaeaacaaIYaaa % aiaadMgaaaa!44D5! = \left( {1 + 2i} \right)\left( {2 - \frac{1}{2}i} \right) - \frac{3}{2}i\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0JaaG % 4maiabgUcaRiaaikdacaWGPbaaaa!3A43! = 3 + 2i\) tọa độ điểm biểu diễn số phức w là: M( 3 ; 2)

Phương trình mặt phẳng (P) theo đoạn chắn: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4baabaGaeyOeI0IaaGOmaaaacqGHRaWkdaWcaaqaaiaadMhaaeaa % caaIZaaaaiabgUcaRmaalaaabaGaamOEaaqaaiabgkHiTiaaiodaaa % Gaeyypa0JaaGymaiabgsDiBlabgkHiTiaaiodacaWG4bGaey4kaSIa % aGOmaiaadMhacqGHsislcaaIYaGaamOEaiabgkHiTiaaiAdacqGH9a % qpcaaIWaaaaa!4E67! \frac{x}{{ - 2}} + \frac{y}{3} + \frac{z}{{ - 3}} = 1 \Leftrightarrow - 3x + 2y - 2z - 6 = 0\)

Dễ thấy mặt phẳng (P) vuông góc với mặt phẳng có phương trình 2x + 2y - z -1 =0 vì tích vô hướng của hai vec-tơ pháp tuyến bằng 0.

Từ  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgU % caRiaaikdacaWGPbGaeyypa0JaaG4maiabgUcaRiaaisdacaWG5bGa % amyAaaaa!3ECC! x + 2i = 3 + 4yi\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aai % qaaqaabeqaaiaadIhacqGH9aqpcaaIZaaabaGaaGOmaiabg2da9iaa % isdacaWG5baaaiaawUhaaaaa!3FB0! \Rightarrow \left\{ \begin{array}{l} x = 3\\ 2 = 4y \end{array} \right.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaadIhacqGH9aqpcaaIZaaabaGaamyEaiabg2da9maa % laaabaGaaGymaaqaaiaaikdaaaaaaiaawUhaaaaa!3FBC! \Leftrightarrow \left\{ \begin{array}{l} x = 3\\ y = \frac{1}{2} \end{array} \right.\)

Vậy , x = 3; \(y= \frac{1}{2}\)

Mặt cầu S có tâm I ( 4 ; 3; -2)  và bán kính R =5.

Gọi H là trung điểm của AB thì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiaadI % eacqGHLkIxcaWGbbGaamOqaaaa!3ACD! IH \bot AB\) và IH = 3 nên H thuộc mặt cầu (S') tâm I bán kính R' = 3 .

Gọi M là trung điểm của A'B' thì AA' + BB' = 2HM,  M nằm trên mặt phẳng (P).

Mặt khác ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamysaiaacUdadaqadaqaaiaadcfaaiaawIcacaGLPaaaaiaa % wIcacaGLPaaacqGH9aqpdaWcaaqaaiaaisdaaeaadaGcaaqaaiaaio % daaSqabaaaaOGaeyipaWJaamOuaaaa!40E2! d\left( {I;\left( P \right)} \right) = \frac{4}{{\sqrt 3 }} < R\) nên (P) cắt mặt cầu (S) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaci4CaiaacM % gacaGGUbWaaeWaaeaacaWGKbGaai4oamaabmaabaGaamiuaaGaayjk % aiaawMcaaaGaayjkaiaawMcaaiabg2da9iGacohacaGGPbGaaiOBai % abeg7aHjabg2da9maalaaabaGaaGynaaqaaiaaiodadaGcaaqaaiaa % iodaaSqabaaaaaaa!4742! \sin \left( {d;\left( P \right)} \right) = \sin \alpha = \frac{5}{{3\sqrt 3 }}\) . Gọi K là hình chiếu của H lên (P) thì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamisaiaadU % eacqGH9aqpcaWGibGaamytaiaac6caciGGZbGaaiyAaiaac6gacqaH % Xoqyaaa!3F5F! HK = HM.\sin \alpha \) .

Vậy để AA' + BB' lớn nhất thì HK lớn nhất

HK đi qua I nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamisaiaadU % eadaWgaaWcbaGaciyBaiaacggacaGG4baabeaakiabg2da9iqadkfa % gaqbaiabgUcaRiaadsgadaqadaqaaiaadMeacaGG7aWaaeWaaeaaca % WGqbaacaGLOaGaayzkaaaacaGLOaGaayzkaaGaeyypa0JaaG4maiab % gUcaRmaalaaabaGaaGinaaqaamaakaaabaGaaG4maaWcbeaaaaGccq % GH9aqpdaWcaaqaaiaaisdacqGHRaWkcaaIZaWaaOaaaeaacaaIZaaa % leqaaaGcbaWaaOaaaeaacaaIZaaaleqaaaaaaaa!4D44! H{K_{\max }} = R' + d\left( {I;\left( P \right)} \right) = 3 + \frac{4}{{\sqrt 3 }} = \frac{{4 + 3\sqrt 3 }}{{\sqrt 3 }}\)

Vậy AA' + BB' lớn nhất bằng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmamaabm % aabaWaaSaaaeaacaaI0aGaey4kaSIaaG4mamaakaaabaGaaG4maaWc % beaaaOqaamaakaaabaGaaG4maaWcbeaaaaaakiaawIcacaGLPaaaca % GGUaWaaSaaaeaacaaIZaWaaOaaaeaacaaIZaaaleqaaaGcbaGaaGyn % aaaacqGH9aqpdaWcaaqaaiaaikdacaaI0aGaey4kaSIaaGymaiaaiI % dadaGcaaqaaiaaiodaaSqabaaakeaacaaI1aaaaaaa!461A! 2\left( {\frac{{4 + 3\sqrt 3 }}{{\sqrt 3 }}} \right).\frac{{3\sqrt 3 }}{5} = \frac{{24 + 18\sqrt 3 }}{5}\)

* Do \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaadg % eacqGHLkIxdaqadaqaaiaadgeacaWGcbGaam4qaiaadseaaiaawIca % caGLPaaaaaa!3DEA! SA \bot \left( {ABCD} \right)\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % 4uaiaadgeacqGHLkIxcaWGbbGaam4qaaaa!3D2E! \Rightarrow SA \bot AC\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aae % caaeaacaWGtbGaamyqaiaadoeaaiaawkWaaiabg2da9iaaiMdacaaI % WaGaeyiSaalaaa!3FE8! \Rightarrow \widehat {SAC} = 90^\circ \)

* Do \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOqaiaado % eacqGHLkIxdaqadaqaaiaadofacaWGbbGaamOqaaGaayjkaiaawMca % aaaa!3D22! BC \bot \left( {SAB} \right)\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OqaiaadoeacqGHLkIxcaWGtbGaam4qaaaa!3D31! \Rightarrow BC \bot SC\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aae % caaeaacaWGtbGaamOqaiaadoeaaiaawkWaaiabg2da9iaaiMdacaaI % WaGaeyiSaalaaa!3FE9! \Rightarrow \widehat {SBC} = 90^\circ \)

* Do \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qaiaadw % eacaqGVaGaae4laiaadgeacaWGcbGaeyO0H4Taam4qaiaadweacqGH % LkIxdaqadaqaaiaadofacaWGbbGaamiraaGaayjkaiaawMcaaaaa!4407! CE{\rm{//}}AB \Rightarrow CE \bot \left( {SAD} \right)\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % 4qaiaadweacqGHLkIxcaWGtbGaamyraaaa!3D36! \Rightarrow CE \bot SE\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aae % caaeaacaWGtbGaamyraiaadoeaaiaawkWaaiabg2da9iaaiMdacaaI % WaGaeyiSaalaaa!3FEC! \Rightarrow \widehat {SEC} = 90^\circ \)

Suy ra các điểm A, B,E cùng nhìn đoạn SC dưới một góc vuông nên mặt cầu đi qua các điểm S, A, C, B, E là mặt cầu đường kính SC

Bán kính mặt cầu đi qua các điểm S, A , C, B, E là: \(R = \frac{{SC}}{2}\)

Xét tam giác SAC vuông tại A  ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaado % eacqGH9aqpcaWGbbGaamOqamaakaaabaGaaGOmaaWcbeaakiabg2da % 9iaadggadaGcaaqaaiaaikdaaSqabaaaaa!3DB9! AC = AB\sqrt 2 = a\sqrt 2 \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % 4uaiaadoeacqGH9aqpcaWGbbGaam4qamaakaaabaGaaGOmaaWcbeaa % kiabg2da9iaaikdacaWGHbaaaa!400E! \Rightarrow SC = AC\sqrt 2 = 2a\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % Ouaiabg2da9maalaaabaGaam4uaiaadoeaaeaacaaIYaaaaiabg2da % 9iaadggaaaa!3E86! \Rightarrow R = \frac{{SC}}{2} = a\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpCpC0xbbL8-4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4saiabg2 % da9maapehabaWaaeWaaeaacaqGLbWaaWbaaSqabeaacaaIXaGaey4k % aSIaciiBaiaac6gadaqadaqaaiaadAgadaqadaqaaiaadIhaaiaawI % cacaGLPaaaaiaawIcacaGLPaaaaaGccqGHRaWkcaaI0aaacaGLOaGa % ayzkaaGaaeizaiaadIhaaSqaaiaaicdaaeaacaaIZaaaniabgUIiYd % GccqGH9aqpdaWdXbqaaiaabwgadaahaaWcbeqaaiaaigdacqGHRaWk % ciGGSbGaaiOBamaabmaabaGaamOzamaabmaabaGaamiEaaGaayjkai % aawMcaaaGaayjkaiaawMcaaaaakiaabsgacaWG4baaleaacaaIWaaa % baGaaG4maaqdcqGHRiI8aOGaey4kaSYaa8qCaeaacaaI0aGaaeizai % aadIhaaSqaaiaaicdaaeaacaaIZaaaniabgUIiYdGccqGH9aqpcaqG % LbGaaiOlamaapehabaGaamOzamaabmaabaGaamiEaaGaayjkaiaawM % caaiaabsgacaWG4baaleaacaaIWaaabaGaaG4maaqdcqGHRiI8aOGa % ey4kaSYaa8qCaeaacaaI0aGaaeizaiaadIhaaSqaaiaaicdaaeaaca % aIZaaaniabgUIiYdGccqGH9aqpcaaI0aGaaeyzaiabgUcaRiaaisda % caWG4bWaaubmaeqaleaacaaIWaaabaGaaG4maaqdbaGaaiiFaaaaki % abg2da9iaaisdacaqGLbGaey4kaSIaaGymaiaaikdaaaa!8337! K = \int\limits_0^3 {\left( {{{\rm{e}}^{1 + \ln \left( {f\left( x \right)} \right)}} + 4} \right){\rm{d}}x} = \int\limits_0^3 {{{\rm{e}}^{1 + \ln \left( {f\left( x \right)} \right)}}{\rm{d}}x} + \int\limits_0^3 {4{\rm{d}}x} = {\rm{e}}.\int\limits_0^3 {f\left( x \right){\rm{d}}x} + \int\limits_0^3 {4{\rm{d}}x} = 4{\rm{e}} + 4x\mathop |\nolimits_0^3 = 4{\rm{e}} + 12\)

Vậy K = 4e + 12

Ta có : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaamaalaaabaWaaOaaaeaacaWG5baameqaaaWc % baGaamiEaaaaaeqaaOWaaSaaaeaadaGcaaqaaiaadMhaaSqabaaake % aadaGcaaqaaiaadIhaaSqabaaaaOGaeyypa0ZaaSaaaeaacaaIXaaa % baGaaGOmaaaadaqadaqaaiGacYgacaGGVbGaai4zamaaBaaaleaada % WcaaqaamaakaaabaGaamyEaaadbeaaaSqaaiaadIhaaaaabeaakmaa % laaabaGaamyEaaqaaiaadIhaaaaacaGLOaGaayzkaaaaaa!48DA! {\log _{\frac{{\sqrt y }}{x}}}\frac{{\sqrt y }}{{\sqrt x }} = \frac{1}{2}\left( {{{\log }_{\frac{{\sqrt y }}{x}}}\frac{y}{x}} \right)\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaaIXaaabaGaaGOmaaaacaGGUaWaaSaaaeaaciGGSbGaai4B % aiaacEgadaWgaaWcbaGaamiEaaqabaGccaWG5bGaeyOeI0IaaGymaa % qaamaalaaabaGaaGymaaqaaiaaikdaaaGaciiBaiaac+gacaGGNbWa % aSbaaSqaaiaadIhaaeqaaOGaamyEaiabgkHiTiaaigdaaaaaaa!481C! = \frac{1}{2}.\frac{{{{\log }_x}y - 1}}{{\frac{1}{2}{{\log }_x}y - 1}}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaaciGGSbGaai4BaiaacEgadaWgaaWcbaGaamiEaaqabaGccaWG % 5bGaeyOeI0IaaGymaaqaaiGacYgacaGGVbGaai4zamaaBaaaleaaca % WG4baabeaakiaadMhacqGHsislcaaIYaaaaaaa!445D! = \frac{{{{\log }_x}y - 1}}{{{{\log }_x}y - 2}}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaaIYaGaciiBaiaac+gacaGGNbWaaSbaaSqaaiaadIhaaeqa % aOWaaOaaaeaacaWG5baaleqaaOGaeyOeI0IaaGymaaqaaiaaikdaci % GGSbGaai4BaiaacEgadaWgaaWcbaGaamiEaaqabaGcdaGcaaqaaiaa % dMhaaSqabaGccqGHsislcaaIYaaaaaaa!461F! = \frac{{2{{\log }_x}\sqrt y - 1}}{{2{{\log }_x}\sqrt y - 2}}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 % da9maabmaabaGaaGOmaiGacYgacaGGVbGaai4zamaaBaaaleaacaWG % 4baabeaakmaakaaabaGaamyEaaWcbeaakiabgkHiTiaaigdaaiaawI % cacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI4aWaaeWa % aeaadaWcaaqaaiaaikdaciGGSbGaai4BaiaacEgadaWgaaWcbaGaam % iEaaqabaGcdaGcaaqaaiaadMhaaSqabaGccqGHsislcaaIXaaabaGa % aGOmaiGacYgacaGGVbGaai4zamaaBaaaleaacaWG4baabeaakmaaka % aabaGaamyEaaWcbeaakiabgkHiTiaaikdaaaaacaGLOaGaayzkaaWa % aWbaaSqabeaacaaIYaaaaaaa!5510! P = {\left( {2{{\log }_x}\sqrt y - 1} \right)^2} + 8{\left( {\frac{{2{{\log }_x}\sqrt y - 1}}{{2{{\log }_x}\sqrt y - 2}}} \right)^2}\)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaaikdaciGGSbGaai4BaiaacEgadaWgaaWcbaGaamiEaaqabaGc % daGcaaqaaiaadMhaaSqabaaaaa!3DCB! t = 2{\log _x}\sqrt y \) do \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiabgY % da8iaadIhacqGH8aapdaGcaaqaaiaadMhaaSqabaGccqGHuhY2ciGG % SbGaai4BaiaacEgadaWgaaWcbaGaamiEaaqabaGccaaIXaGaeyipaW % JaciiBaiaac+gacaGGNbWaaSbaaSqaaiaadIhaaeqaaOGaamiEaiab % gYda8iGacYgacaGGVbGaai4zamaaBaaaleaacaWG4baabeaakmaaka % aabaGaamyEaaWcbeaaaaa!4E15! 1 < x < \sqrt y \Leftrightarrow {\log _x}1 < {\log _x}x < {\log _x}\sqrt y \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % iDaiabg6da+iaaikdaaaa!3B0E! \Rightarrow t > 2\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG0baacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaaI % YaWaaeWaaeaacaWG0bGaeyOeI0IaaGymaaGaayjkaiaawMcaamaabm % aabaGaamiDaiabgkHiTiaaisdaaiaawIcacaGLPaaadaqadaqaaiaa % dshadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaIYaGaamiDaiabgU % caRiaaisdaaiaawIcacaGLPaaaaeaadaqadaqaaiaadshacqGHsisl % caaIYaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIZaaaaaaaaaa!5062! f'\left( t \right) = \frac{{2\left( {t - 1} \right)\left( {t - 4} \right)\left( {{t^2} - 2t + 4} \right)}}{{{{\left( {t - 2} \right)}^3}}}\) ; \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG0baacaGLOaGaayzkaaGaeyypa0JaaGimaiabgkDi % EpaadeaaeaqabeaacaWG0bGaeyypa0JaaGymaaqaaiaadshacqGH9a % qpcaaI0aaaaiaawUfaaaaa!4403! f'\left( t \right) = 0 \Rightarrow \left[ \begin{array}{l} t = 1\\ t = 4 \end{array} \right.\)

Lập bảng biến thiên trên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIYaGaai4oaiabgUcaRiabg6HiLcGaayjkaiaawMcaaaaa!3B4B! \left( {2; + \infty } \right)\) ta được

Vậy giá trị nhỏ nhất của biểu thức \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 % da9maabmaabaGaciiBaiaac+gacaGGNbWaaSbaaSqaaiaadIhaaeqa % aOGaamyEaiabgkHiTiaaigdaaiaawIcacaGLPaaadaahaaWcbeqaai % aaikdaaaGccqGHRaWkcaaI4aWaaeWaaeaaciGGSbGaai4BaiaacEga % daWgaaWcbaWaaSaaaeaadaGcaaqaaiaadMhaaWqabaaaleaacaWG4b % aaaaqabaGcdaWcaaqaamaakaaabaGaamyEaaWcbeaaaOqaamaakaaa % baGaamiEaaWcbeaaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaik % daaaaaaa!4C97! P = {\left( {{{\log }_x}y - 1} \right)^2} + 8{\left( {{{\log }_{\frac{{\sqrt y }}{x}}}\frac{{\sqrt y }}{{\sqrt x }}} \right)^2}\) là  27 đạt được khi  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaaisdacqGHuhY2caaIYaGaciiBaiaac+gacaGGNbWaaSbaaSqa % aiaadIhaaeqaaOWaaOaaaeaacaWG5baaleqaaOGaeyypa0JaaGinaa % aa!42B3! t = 4 \Leftrightarrow 2{\log _x}\sqrt y = 4\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aaO % aaaeaacaWG5baaleqaaOGaeyypa0JaamiEamaaCaaaleqabaGaaGOm % aaaakiabgsDiBlaadMhacqGH9aqpcaWG4bWaaWbaaSqabeaacaaI0a % aaaaaa!42B1! \Leftrightarrow \sqrt y = {x^2} \Leftrightarrow y = {x^4}\)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadAgadaqadaqaaiaa % dIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI4aGaamiEaiabgU % caRiaad2gaaiaawIcacaGLPaaaaaa!434F! g\left( x \right) = f\left( {{x^2} - 8x + m} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0ZaaeWaaeaacaWG % 4bGaeyOeI0IaaGymaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaa % aakmaabmaabaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaa % ikdacaWG4baacaGLOaGaayzkaaGaeyO0H4naaa!4813! f'\left( x \right) = {\left( {x - 1} \right)^2}\left( {{x^2} - 2x} \right) \Rightarrow \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4zayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0ZaaeWaaeaacaaI % YaGaamiEaiabgkHiTiaaiIdaaiaawIcacaGLPaaadaqadaqaaiaadI % hadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI4aGaamiEaiabgUca % Riaad2gacqGHsislcaaIXaaacaGLOaGaayzkaaWaaWbaaSqabeaaca % aIYaaaaOWaaeWaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOe % I0IaaGioaiaadIhacqGHRaWkcaWGTbaacaGLOaGaayzkaaWaaeWaae % aacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGioaiaadIha % cqGHRaWkcaWGTbGaeyOeI0IaaGOmaaGaayjkaiaawMcaaaaa!5B97! g'\left( x \right) = \left( {2x - 8} \right){\left( {{x^2} - 8x + m - 1} \right)^2}\left( {{x^2} - 8x + m} \right)\left( {{x^2} - 8x + m - 2} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4zayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaaGimaaaa!3B31! g'\left( x \right) = 0\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aam % qaaqaabeqaaiaadIhacqGH9aqpcaaI0aaabaGaamiEamaaCaaaleqa % baGaaGOmaaaakiabgkHiTiaaiIdacaWG4bGaey4kaSIaamyBaiabgk % HiTiaaigdacqGH9aqpcaaIWaGaaGPaVlaaykW7caaMc8UaaGPaVlaa % ykW7daqadaqaaiaaigdaaiaawIcacaGLPaaaaeaacaWG4bWaaWbaaS % qabeaacaaIYaaaaOGaeyOeI0IaaGioaiaadIhacqGHRaWkcaWGTbGa % eyypa0JaaGimaiaaykW7caaMc8UaaGPaVlaaykW7caaMc8UaaGPaVl % aaykW7caaMc8UaaGPaVlaaykW7caaMc8+aaeWaaeaacaaIYaaacaGL % OaGaayzkaaaabaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTi % aaiIdacaWG4bGaey4kaSIaamyBaiabgkHiTiaaikdacqGH9aqpcaaI % WaGaaGPaVlaaykW7caaMc8+aaeWaaeaacaaIZaaacaGLOaGaayzkaa % aaaiaawUfaaaaa!7C15! \Leftrightarrow \left[ \begin{array}{l} x = 4\\ {x^2} - 8x + m - 1 = 0\,\,\,\,\,\left( 1 \right)\\ {x^2} - 8x + m = 0\,\,\,\,\,\,\,\,\,\,\,\left( 2 \right)\\ {x^2} - 8x + m - 2 = 0\,\,\,\left( 3 \right) \end{array} \right.\)

Các phương trình (1),(2) ,(3)  không có nghiệm chung từng đôi một và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGioaiaadIhacqGH % RaWkcaWGTbGaeyOeI0IaaGymaaGaayjkaiaawMcaamaaCaaaleqaba % GaaGOmaaaakiabgwMiZkaaicdaaaa!4307! {\left( {{x^2} - 8x + m - 1} \right)^2} \ge 0\) với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiaIiIaam % iEaiabgIGiolabl2riHcaa!3AB4! \forall x \in R\)

Suy ra g(x) có 5 điểm cực trị khi và chỉ khi (2) và (3) có hai nghiệm phân biệt khác 4.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiabfs5aenaaBaaaleaacaaIYaaabeaakiabg2da9iaa % igdacaaI2aGaeyOeI0IaamyBaiabg6da+iaaicdaaeaacqqHuoarda % WgaaWcbaGaaG4maaqabaGccqGH9aqpcaaIXaGaaGOnaiabgkHiTiaa % d2gacqGHRaWkcaaIYaGaeyOpa4JaaGimaaqaaiaaigdacaaI2aGaey % OeI0IaaG4maiaaikdacqGHRaWkcaWGTbGaeyiyIKRaaGimaaqaaiaa % igdacaaI2aGaeyOeI0IaaG4maiaaikdacqGHRaWkcaWGTbGaeyOeI0 % IaaGOmaiabgcMi5kaaicdaaaGaay5Eaaaaaa!5E1A! \Leftrightarrow \left\{ \begin{array}{l} {\Delta _2} = 16 - m > 0\\ {\Delta _3} = 16 - m + 2 > 0\\ 16 - 32 + m \ne 0\\ 16 - 32 + m - 2 \ne 0 \end{array} \right.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaad2gacqGH8aapcaaIXaGaaGOnaaqaaiaad2gacqGH % 8aapcaaIXaGaaGioaaqaaiaad2gacqGHGjsUcaaIXaGaaGOnaaqaai % aad2gacqGHGjsUcaaIXaGaaGioaaaacaGL7baaaaa!48C0! \Leftrightarrow \left\{ \begin{array}{l} m < 16\\ m < 18\\ m \ne 16\\ m \ne 18 \end{array} \right.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % yBaiabgYda8iaaigdacaaI2aaaaa!3BC0! \Leftrightarrow m < 16\)

Vì m nguyên dương và m < 16 nên có 15 giá trị m cần tìm.

Số tập con gồm 2 phần tử của M là số cách chọn 2 phần tử bất kì trong 10 phần tử của M. Do đó số tập con gồm 2 phần tử của M là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qamaaDa % aaleaacaaIXaGaaGimaaqaaiaaikdaaaaaaa!391A! C_{10}^2\).

Ta có tứ giác BOKC là tứ giác nội tiếp đường tròn suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGpbGaam4saiaadkeaaiaawkWaaiabg2da9maaHaaabaGaam4taiaa % doeacaWGcbaacaGLcmaaaaa!3D4C! \widehat {OKB} = \widehat {OCB}\) (1)

Ta có tứ giác KDHC là tứ giác nội tiếp đường tròn suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGebGaam4saiaadIeaaiaawkWaaiabg2da9maaHaaabaGaam4taiaa % doeacaWGcbaacaGLcmaaaaa!3D47! \widehat {DKH} = \widehat {OCB}\)(2)

Từ (1) và (2) suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGebGaam4saiaadIeaaiaawkWaaiabg2da9maaHaaabaGaam4taiaa % dUeacaWGcbaacaGLcmaaaaa!3D4F! \widehat {DKH} = \widehat {OKB}\) .

Do đó BK là đường phân giác trong của góc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGpbGaam4saiaadIeaaiaawkWaaaaa!3927! \widehat {OKH}\) và AC là đường phân giác ngoài của góc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGpbGaam4saiaadIeaaiaawkWaaaaa!3927! \widehat {OKH}\) .

Tương tự ta chứng minh được OC là đường phân giác trong của góc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGlbGaam4taiaadIeaaiaawkWaaaaa!3927! \widehat {KOH}\) và AB là đường phân giác ngoài của góc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGlbGaam4taiaadIeaaiaawkWaaaaa!3927! \widehat {KOH}\).

Ta có OK = 4; OH = 3; KH = 5

Gọi I, J lần lượt là chân đường phân giác ngoài của góc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGpbGaam4saiaadIeaaiaawkWaaaaa!3927! \widehat {OKH}\) và \(\widehat {KOH}\).

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9iaadgeacaWGdbGaeyykICSaamisaiaad+eaaaa!3C95! I = AC \cap HO\) ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGjbGaam4taaqaaiaadMeacaWGibaaaiabg2da9maalaaabaGaam4s % aiaad+eaaeaacaWGlbGaamisaaaacqGH9aqpdaWcaaqaaiaaisdaae % aacaaI1aaaaaaa!402B! \frac{{IO}}{{IH}} = \frac{{KO}}{{KH}} = \frac{4}{5}\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aa8 % HaaeaacaWGjbGaam4taaGaay51GaGaeyypa0ZaaSaaaeaacaaI0aaa % baGaaGynaaaadaWhcaqaaiaadMeacaWGibaacaGLxdcaaaa!4189! \Rightarrow \overrightarrow {IO} = \frac{4}{5}\overrightarrow {IH} \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % ysamaabmaabaGaeyOeI0IaaGioaiaacUdacaaMc8UaeyOeI0IaaGio % aiaacUdacaaMc8UaeyOeI0IaaGinaaGaayjkaiaawMcaaaaa!4445! \Rightarrow I\left( { - 8;\, - 8;\, - 4} \right)\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOsaiabg2 % da9iaadgeacaWGcbGaeyykICSaam4saiaadIeaaaa!3C91! J = AB \cap KH\) ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGkbGaam4saaqaaiaadQeacaWGibaaaiabg2da9maalaaabaGaam4t % aiaadUeaaeaacaWGpbGaamisaaaacqGH9aqpdaWcaaqaaiaaisdaae % aacaaIZaaaaaaa!402B! \frac{{JK}}{{JH}} = \frac{{OK}}{{OH}} = \frac{4}{3}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H49aa8 % HaaeaacaWGkbGaam4saaGaay51GaGaeyypa0ZaaSaaaeaacaaI0aaa % baGaaG4maaaadaWhcaqaaiaadQeacaWGibaacaGLxdcacqGHshI3ca % WGkbWaaeWaaeaacaaIXaGaaGOnaiaacUdacaaMc8UaaGinaiaacUda % caaMc8UaeyOeI0IaaGinaaGaayjkaiaawMcaaaaa!4EB2! \Rightarrow \overrightarrow {JK} = \frac{4}{3}\overrightarrow {JH} \Rightarrow J\left( {16;\,4;\, - 4} \right)\).

Đường thẳng IK qua I nhận \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGjbGaam4saaGaay51GaGaeyypa0ZaaeWaaeaadaWcaaqaaiaaigda % caaI2aaabaGaaG4maaaacaGG7aGaaGPaVpaalaaabaGaaGOmaiaaiI % daaeaacaaIZaaaaiaacUdacaaMc8+aaSaaaeaacaaIYaGaaGimaaqa % aiaaiodaaaaacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaaI0aaaba % GaaG4maaaadaqadaqaaiaaisdacaGG7aGaaGPaVlaaiEdacaGG7aGa % aGPaVlaaiwdaaiaawIcacaGLPaaaaaa!522B! \overrightarrow {IK} = \left( {\frac{{16}}{3};\,\frac{{28}}{3};\,\frac{{20}}{3}} \right) = \frac{4}{3}\left( {4;\,7;\,5} \right)\) làm vec tơ chỉ phương có phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGjbGaam4saaGaayjkaiaawMcaaiaacQdadaGabaabaeqabaGaamiE % aiabg2da9iabgkHiTiaaiIdacqGHRaWkcaaI0aGaamiDaaqaaiaadM % hacqGH9aqpcqGHsislcaaI4aGaey4kaSIaaG4naiaadshaaeaacaWG % 6bGaeyypa0JaeyOeI0IaaGinaiabgUcaRiaaiwdacaWG0baaaiaawU % haaaaa!4DDF! \left( {IK} \right):\left\{ \begin{array}{l} x = - 8 + 4t\\ y = - 8 + 7t\\ z = - 4 + 5t \end{array} \right.\).

Đường thẳng OJ qua O nhận \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGpbGaamOsaaGaay51GaGaeyypa0ZaaeWaaeaacaaIXaGaaGOnaiaa % cUdacaaMc8UaaGinaiaacUdacaaMc8UaeyOeI0IaaGinaaGaayjkai % aawMcaaiabg2da9iaaisdadaqadaqaaiaaisdacaGG7aGaaGPaVlaa % igdacaGG7aGaaGPaVlabgkHiTiaaigdaaiaawIcacaGLPaaaaaa!4F54! \overrightarrow {OJ} = \left( {16;\,4;\, - 4} \right) = 4\left( {4;\,1;\, - 1} \right)\) làm vec tơ chỉ phương có phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGpbGaamOsaaGaayjkaiaawMcaaiaacQdadaGabaabaeqabaGaamiE % aiabg2da9iaaisdaceWG0bGbauaaaeaacaWG5bGaeyypa0JabmiDay % aafaaabaGaamOEaiabg2da9iabgkHiTiqadshagaqbaaaacaGL7baa % aaa!45C6! \left( {OJ} \right):\left\{ \begin{array}{l} x = 4t'\\ y = t'\\ z = - t' \end{array} \right.\).

Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabg2 % da9iaadMeacaWGlbGaeyykICSaam4taiaadQeaaaa!3C9F! A = IK \cap OJ\), giải hệ ta tìm được A(-4 ; -1; 1).

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGjbGaamyqaaGaay51GaGaeyypa0ZaaeWaaeaacaaI0aGaai4oaiaa % ykW7caaI3aGaai4oaiaaykW7caaI1aaacaGLOaGaayzkaaaaaa!429D! \overrightarrow {IA} = \left( {4;\,7;\,5} \right)\) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGjbGaamOsaaGaay51GaGaeyypa0ZaaeWaaeaacaaIYaGaaGinaiaa % cUdacaaMc8UaaGymaiaaikdacaGG7aGaaGPaVlaaicdaaiaawIcaca % GLPaaaaaa!4413! \overrightarrow {IJ} = \left( {24;\,12;\,0} \right)\) , ta tính \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamWaaeaada % WhcaqaaiaadMeacaWGbbaacaGLxdcacaGGSaWaa8HaaeaacaWGjbGa % amOsaaGaay51GaaacaGLBbGaayzxaaGaeyypa0ZaaeWaaeaacqGHsi % slcaaI2aGaaGimaiaacUdacaaIXaGaaGOmaiaaicdacaGG7aGaeyOe % I0IaaGymaiaaikdacaaIWaaacaGLOaGaayzkaaGaeyypa0JaeyOeI0 % IaaGOnaiaaicdadaqadaqaaiaaigdacaGG7aGaaGPaVlabgkHiTiaa % ikdacaGG7aGaaGPaVlaaikdaaiaawIcacaGLPaaaaaa!579C! \left[ {\overrightarrow {IA} ,\overrightarrow {IJ} } \right] = \left( { - 60;120; - 120} \right) = - 60\left( {1;\, - 2;\,2} \right)\).

Khi đó đường thẳng đi qua A và vuông góc với mặt phẳng (ABC) có véc tơ chỉ phương \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WG1baacaGLxdcacqGH9aqpdaqadaqaaiaaigdacaGG7aGaeyOeI0Ia % aGOmaiaacUdacaaIYaaacaGLOaGaayzkaaaaaa!3FCF! \overrightarrow u = \left( {1; - 2;2} \right)\) nên có phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4bGaey4kaSIaaGinaaqaaiaaigdaaaGaeyypa0ZaaSaaaeaacaWG % 5bGaey4kaSIaaGymaaqaaiabgkHiTiaaikdaaaGaeyypa0ZaaSaaae % aacaWG6bGaeyOeI0IaaGymaaqaaiaaikdaaaaaaa!432F! \frac{{x + 4}}{1} = \frac{{y + 1}}{{ - 2}} = \frac{{z - 1}}{2}\).

Chọn hệ tọa độ Oxy , khi đó : 

Diện tích hình chữ nhật là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaaIXaaabeaakiabg2da9iaaisdacqaHapaCaaa!3B3D! {S_1} = 4\pi \).

Diện tích phần đất được tô màu đen là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaaIYaaabeaakiabg2da9iaaikdadaWdXbqaaiGacohacaGG % PbGaaiOBaiaadIhaaSqaaiaaicdaaeaacqaHapaCa0Gaey4kIipaki % aabsgacaWG4bGaeyypa0JaaGinaaaa!45E6! {S_2} = 2\int\limits_0^\pi {\sin x} {\rm{d}}x = 4\).

Tính diện tích phần còn lại: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9iaadofadaWgaaWcbaGaaGymaaqabaGccqGHsislcaWGtbWaaSba % aSqaaiaaikdaaeqaaOGaeyypa0JaaGinaiabec8aWjabgkHiTiaais % dacqGH9aqpcaaI0aWaaeWaaeaacqaHapaCcqGHsislcaaIXaaacaGL % OaGaayzkaaaaaa!482F! S = {S_1} - {S_2} = 4\pi - 4 = 4\left( {\pi - 1} \right)\)

Ta có: A(2; 2;2) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiaadg % eacqGH9aqpcaWGqbGaamOqaiabg2da9iaadcfacaWGdbGaeyypa0Za % aSaaaeaacaaIZaWaaOaaaeaacaaIYaGaaGymaaWcbeaaaOqaaiaais % daaaaaaa!4101! PA = PB = PC = \frac{{3\sqrt {21} }}{4}\).

Trong \(\Delta OBC\) kẻ \(OI \perp BC\)

Từ đề bài ta có \( \left\{ \begin{array}{l} OA \bot OB\\ OA \bot OC \end{array} \right.\)\(\Rightarrow OA \perp BC\)

\( \left\{ \begin{array}{l} BC \bot OI\\ BC \bot OA \end{array} \right.\)\(\Rightarrow BC \perp AI\)

Ta có: \( \left\{ \begin{array}{l} \left( {OBC} \right) \cap \left( {ABC} \right) = BC\\ BC \bot AI\\ BC \bot OI \end{array} \right. \Rightarrow \widehat {\left( {\left( {OBC} \right),\left( {ABC} \right)} \right)} = \widehat {\left( {OI,AI} \right)} = \widehat {OIA}\)

Ta có: \(OI = \frac{1}{2}BC = \frac{1}{2}\sqrt {O{B^2} + O{C^2}} = a\sqrt 3 \)

Xét tam giác OAI vuông tại A có \(\tan \widehat {OIA} = \frac{{OA}}{{OI}} = \frac{{\sqrt 3 }}{3} \Rightarrow \widehat {OIA} = 30^\circ \).

Vậy \(\widehat {\left( {\left( {OBC} \right),\left( {ABC} \right)} \right)} = 30^\circ \).

Ta có tập xác định: D =R\ {1}

Do \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcqGHXcqScqGHEisP % aeqaaOGaamyEaiabg2da9iaaiodaaaa!4210! \mathop {\lim }\limits_{x \to \pm \infty } y = 3\) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcaaIXaWaaWbaaWqa % beaacqGHRaWkaaaaleqaaOGaamyEaiabg2da9iabgkHiTiabg6HiLc % aa!4228! \mathop {\lim }\limits_{x \to {1^ + }} y = - \infty \) , \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcaaIXaWaaWbaaWqa % beaacqGHsislaaaaleqaaOGaamyEaiabg2da9iabgUcaRiabg6HiLc % aa!4228! \mathop {\lim }\limits_{x \to {1^ - }} y = + \infty \) nên đồ thị hàm số có hai đường tiệm cận.

Do \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiabgw % QiEnaabmaabaGaamiuaaGaayjkaiaawMcaaaaa!3AEC! d \bot \left( P \right)\) nên vec-tơ chỉ phương của đường thẳng d là vec-tơ pháp tuyến của (P) .

Suy ra một một vec-tơ chỉ phương của đường thẳng d là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WG1baacaGLxdcacqGH9aqpdaWhcaqaaiaad6gadaWgaaWcbaWaaeWa % aeaacaWGqbaacaGLOaGaayzkaaaabeaaaOGaay51GaGaeyypa0Zaae % WaaeaacaaI0aGaai4oaiaaysW7caaIWaGaai4oaiaaysW7cqGHsisl % caaIXaaacaGLOaGaayzkaaaaaa!492A! \overrightarrow u = \overrightarrow {{n_{\left( P \right)}}} = \left( {4;\;0;\; - 1} \right)\) .

Gọi A ( a; 0;0),B ( 0; b; 0)  và C ( 0;0;c) với \( abc \ne 0\).

Phương trình mặt phẳng (P) đi qua ba điểm A,B,C  là \(\frac{x}{a} + \frac{y}{b} + \frac{z}{c} = 1\)

Vì \( M\left( {1\,;\,2\,;\,3} \right) \in \left( P \right)\) nên ta có: \( \frac{1}{a} + \frac{2}{b} + \frac{3}{c} = 1\).

Điểm M là trực tâm của \( \Delta ABC \Leftrightarrow \left\{ \begin{array}{l} AM \bot BC\\ BM \bot AC \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \overrightarrow {AM} \,.\,\overrightarrow {BC} = 0\\ \overrightarrow {BM} \,.\,\overrightarrow {AC} = 0 \end{array} \right.\).

Ta có: \(\overrightarrow {AM} = \left( {1 - a\,;\,2\,;\,3} \right); % MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGcbGaam4qaaGaay51GaGaeyypa0ZaaeWaaeaacaaIWaGaaGPaVlaa % cUdacaaMc8UaeyOeI0IaamOyaiaaykW7caGG7aGaaGPaVlaadogaai % aawIcacaGLPaaaaaa!46E6! \overrightarrow {BC} = \left( {0\,;\, - b\,;\,c} \right)\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGcbGaamytaaGaay51GaGaeyypa0ZaaeWaaeaacaaIXaGaaGPaVlaa % cUdacaaMc8UaaGOmaiabgkHiTiaadkgacaaMc8Uaai4oaiaaykW7ca % aIZaaacaGLOaGaayzkaaaaaa!4782! ;\overrightarrow {BM} = \left( {1\,;\,2 - b\,;\,3} \right)\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGbbGaam4qaaGaay51GaGaeyypa0ZaaeWaaeaacqGHsislcaWGHbGa % aGPaVlaacUdacaaMc8UaaGimaiaaykW7caGG7aGaaGPaVlaadogaai % aawIcacaGLPaaaaaa!46E4! ;\overrightarrow {AC} = \left( { - a\,;\,0\,;\,c} \right)\)

Ta có hệ phương trình:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiabgkHiTiaaikdacaWGIbGaey4kaSIaaG4maiaadogacqGH9aqp % caaIWaaabaGaeyOeI0IaamyyaiabgUcaRiaaiodacaWGJbGaeyypa0 % JaaGimaaqaamaalaaabaGaaGymaaqaaiaadggaaaGaey4kaSYaaSaa % aeaacaaIYaaabaGaamOyaaaacqGHRaWkdaWcaaqaaiaaiodaaeaaca % WGJbaaaiabg2da9iaaigdaaaGaay5EaaGaeyi1HS9aaiqaaqaabeqa % aiaadkgacqGH9aqpdaWcaaqaaiaaiodaaeaacaaIYaaaaiaadogaae % aacaWGHbGaeyypa0JaaG4maiaadogaaeaadaWcaaqaaiaaigdaaeaa % caaIZaGaam4yaaaacqGHRaWkdaWcaaqaaiaaikdaaeaadaWcaaqaai % aaiodaaeaacaaIYaaaaiaadogaaaGaey4kaSYaaSaaaeaacaaIZaaa % baGaam4yaaaacqGH9aqpcaaIXaaaaiaawUhaaaaa!62F9! \left\{ \begin{array}{l} - 2b + 3c = 0\\ - a + 3c = 0\\ \frac{1}{a} + \frac{2}{b} + \frac{3}{c} = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} b = \frac{3}{2}c\\ a = 3c\\ \frac{1}{{3c}} + \frac{2}{{\frac{3}{2}c}} + \frac{3}{c} = 1 \end{array} \right.\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiabgkHiTiaaikdacaWGIbGaey4kaSIaaG4maiaadogacqGH9aqp % caaIWaaabaGaeyOeI0IaamyyaiabgUcaRiaaiodacaWGJbGaeyypa0 % JaaGimaaqaamaalaaabaGaaGymaaqaaiaadggaaaGaey4kaSYaaSaa % aeaacaaIYaaabaGaamOyaaaacqGHRaWkdaWcaaqaaiaaiodaaeaaca % WGJbaaaiabg2da9iaaigdaaaGaay5EaaGaeyi1HS9aaiqaaqaabeqa % aiaadkgacqGH9aqpdaWcaaqaaiaaiodaaeaacaaIYaaaaiaadogaae % aacaWGHbGaeyypa0JaaG4maiaadogaaeaadaWcaaqaaiaaigdaaeaa % caaIZaGaam4yaaaacqGHRaWkdaWcaaqaaiaaikdaaeaadaWcaaqaai % aaiodaaeaacaaIYaaaaiaadogaaaGaey4kaSYaaSaaaeaacaaIZaaa % baGaam4yaaaacqGH9aqpcaaIXaaaaiaawUhaaaaa!62F9! \left\{ \begin{array}{l} - 2b + 3c = 0\\ - a + 3c = 0\\ \frac{1}{a} + \frac{2}{b} + \frac{3}{c} = 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} b = \frac{3}{2}c\\ a = 3c\\ \frac{1}{{3c}} + \frac{2}{{\frac{3}{2}c}} + \frac{3}{c} = 1 \end{array} \right.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaadggacqGH9aqpcaaIXaGaaGinaaqaaiaadkgacqGH % 9aqpcaaI3aaabaGaam4yaiabg2da9maalaaabaGaaGymaiaaisdaae % aacaaIZaaaaaaacaGL7baaaaa!43B9! \Leftrightarrow \left\{ \begin{array}{l} a = 14\\ b = 7\\ c = \frac{{14}}{3} \end{array} \right.\)

Phương trình mặt phẳng (P) là \( \frac{x}{{14}} + \frac{y}{7} + \frac{{3z}}{{14}} = 1\)\(\Leftrightarrow x + 2y + 3z - 14 = 0\)

Điều kiện : 3x -1 > 0 \(\iff x > \frac{1}{3}\)

Ta có \( {\log _2}\left( {3x - 1} \right) > 3 \Leftrightarrow 3x - 1 > 8 \Leftrightarrow x > 3\)( nhận )

Đặt M = x , ( x > 0) và OA = a, ( a > 0 ), a là hằng số.

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGnbGaamOtaaqaaiaadofacaWGpbaaaiabg2da9maalaaabaGaamOt % aiaadgeaaeaacaWGpbGaamyqaaaaaaa!3D9E! \frac{{MN}}{{SO}} = \frac{{NA}}{{OA}}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OtaiaadgeacqGH9aqpdaWcaaqaaiaad2eacaWGobGaaiOlaiaad+ea % caWGbbaabaGaam4uaiaad+eaaaaaaa!409D! \Rightarrow NA = \frac{{MN.OA}}{{SO}}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % OtaiaadgeacqGH9aqpdaWcaaqaaiaadIhacaWGHbaabaGaamiAaaaa % aaa!3DD0! \Rightarrow NA = \frac{{xa}}{h}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % 4taiaad6eacqGH9aqpcaWGHbGaeyOeI0YaaSaaaeaacaWG4bGaamyy % aaqaaiaadIgaaaaaaa!3FB1! \Rightarrow ON = a - \frac{{xa}}{h}\)

Khối trụ thu được có bán kính đáy bằng ON và chiều cao bằng MN .

Thể tích khối trụ là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9iabec8aWjaac6cacaWGpbGaamOtamaaCaaaleqabaGaaGOmaaaa % kiaac6cacaWGnbGaamOtaaaa!3F35! V = \pi .O{N^2}.MN\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Jaeq % iWdaNaaiOlaiaadIhacaGGUaGaamyyamaaCaaaleqabaGaaGOmaaaa % kmaabmaabaWaaSaaaeaacaWGObGaeyOeI0IaamiEaaqaaiaadIgaaa % aacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaaa!4337! = \pi .x.{a^2}{\left( {\frac{{h - x}}{h}} \right)^2}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Jaeq % iWdaNaamyyamaaCaaaleqabaGaaGOmaaaakmaalaaabaGaaGymaaqa % aiaaikdacaWGObWaaWbaaSqabeaacaaIYaaaaaaakiaaikdacaWG4b % WaaeWaaeaacaWGObGaeyOeI0IaamiEaaGaayjkaiaawMcaamaaCaaa % leqabaGaaGOmaaaaaaa!44F9! = \pi {a^2}\frac{1}{{2{h^2}}}2x{\left( {h - x} \right)^2}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyizIm6aaS % aaaeaacqaHapaCcaWGHbWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGOm % aiaadIgadaahaaWcbeqaaiaaikdaaaaaaOWaaeWaaeaadaWcaaqaai % aaikdacaWGObaabaGaaG4maaaaaiaawIcacaGLPaaadaahaaWcbeqa % aiaaiodaaaaaaa!42D4! \le \frac{{\pi {a^2}}}{{2{h^2}}}{\left( {\frac{{2h}}{3}} \right)^3}\)

Dấu bằng xảy ra khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiaadI % hacqGH9aqpcaWGObGaeyOeI0IaamiEaaaa!3B8A! 2x = h - x\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % iEaiabg2da9maalaaabaGaamiAaaqaaiaaiodaaaaaaa!3C0D! \Leftrightarrow x = \frac{h}{3}\)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadwhacqGH9aqpciGGSbGaaiOBamaabmaabaGaamiEamaaCaaa % leqabaGaaGOmaaaakiabgUcaRiaaiMdaaiaawIcacaGLPaaaaeaaca % qGKbGaamODaiabg2da9iaadIhacaqGKbGaamiEaaaacaGL7baacqGH % uhY2daGabaabaeqabaGaaeizaiaadwhacqGH9aqpdaWcaaqaaiaaik % dacaWG4baabaWaaeWaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGa % ey4kaSIaaGyoaaGaayjkaiaawMcaaaaacaqGKbGaamiEaaqaaiaadA % hacqGH9aqpdaWcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGH % RaWkcaaI5aaabaGaaGOmaaaaaaGaay5Eaaaaaa!5B70! \left\{ \begin{array}{l} u = \ln \left( {{x^2} + 9} \right)\\ {\rm{d}}v = x{\rm{d}}x \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {\rm{d}}u = \frac{{2x}}{{\left( {{x^2} + 9} \right)}}{\rm{d}}x\\ v = \frac{{{x^2} + 9}}{2} \end{array} \right.\)

Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaaca % WG4bGaciiBaiaac6gadaqadaqaaiaadIhadaahaaWcbeqaaiaaikda % aaGccqGHRaWkcaaI5aaacaGLOaGaayzkaaGaaeizaiaadIhaaSqaai % aaicdaaeaacaaI0aaaniabgUIiYdGccqGH9aqpdaabcaqaamaalaaa % baGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiMdaaeaaca % aIYaaaaiGacYgacaGGUbWaaeWaaeaacaWG4bWaaWbaaSqabeaacaaI % YaaaaOGaey4kaSIaaGyoaaGaayjkaiaawMcaaaGaayjcSdWaa0baaS % qaaiaaicdaaeaacaaI0aaaaOGaeyOeI0Yaa8qCaeaadaWcaaqaaiaa % dIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI5aaabaGaaGOmaa % aacaGGUaWaaSaaaeaacaaIYaGaamiEaaqaaiaadIhadaahaaWcbeqa % aiaaikdaaaGccqGHRaWkcaaI5aaaaiaabsgacaWG4baaleaacaaIWa % aabaGaaGinaaqdcqGHRiI8aaaa!6492! \int\limits_0^4 {x\ln \left( {{x^2} + 9} \right){\rm{d}}x} = \left. {\frac{{{x^2} + 9}}{2}\ln \left( {{x^2} + 9} \right)} \right|_0^4 - \int\limits_0^4 {\frac{{{x^2} + 9}}{2}.\frac{{2x}}{{{x^2} + 9}}{\rm{d}}x} \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0JaaG % OmaiaaiwdaciGGSbGaaiOBaiaaiwdacqGHsislcaaI5aGaciiBaiaa % c6gacaaIZaGaeyOeI0IaaGioaaaa!4118! = 25\ln 5 - 9\ln 3 - 8\)

Do đó a = 25,b = -9  , c = - 8 nên T =8 .

Diện tích đáy:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaaqaaaaaaaaaWdbiabfs5aejaadgeacaWGcbGaam4qaaWdaeqa % aOGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGOmaaaacaGGUaGaaG4mai % aac6cacaaIZaGaaiOlaiGacohacaGGPbGaaiOBaiaaiAdacaaIWaGa % eyiSaaRaeyypa0ZaaSaaaeaacaaI5aWaaOaaaeaacaaIZaaaleqaaa % GcbaGaaGinaaaaaaa!4AC0! {S_{\Delta ABC}} = \frac{1}{2}.3.3.\sin 60^\circ = \frac{{9\sqrt 3 }}{4}\) . Thể tích \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGSbGaamiDaaqabaGccqGH9aqpcaWGtbWaaSbaaSqaaaba % aaaaaaaapeGaeuiLdqKaamyqaiaadkeacaWGdbaapaqabaGccaGGUa % Gaamyqaiqadgeagaqbaiabg2da9maalaaabaGaaGOmaiaaiEdadaGc % aaqaaiaaiodaaSqabaaakeaacaaI0aaaaaaa!456A! {V_{lt}} = {S_{\Delta ABC}}.AA' = \frac{{27\sqrt 3 }}{4}\).

Ta có: \(y' = 3{x^2} - 6x + m\)

Hàm số đạt cực tiểu tại \( x = 2 \Rightarrow y'\left( 2 \right) = 0 \Leftrightarrow m = 0\)

Thử lại: với m = 0 thì \(y' = 3{x^2} - 6x\) \(\Rightarrow y'' = 6x - 6\) \( \Rightarrow y''\left( 2 \right) = 6 > 0\)  suy ra hàm số đạt cực tiểu tại x = 2.