Ta có: 

\(P = \frac{{{a^{\frac{1}{3}}}\left( {{a^{\frac{1}{2}}} - {a^{\frac{5}{2}}}} \right)}}{{{a^{\frac{1}{4}}}\left( {{a^{\frac{7}{{12}}}} - {a^{\frac{{19}}{{12}}}}} \right)}} = \frac{{{a^{\frac{1}{3}}} \cdot {a^{\frac{1}{2}}}\left( {1 - {a^2}} \right)}}{{{a^{\frac{1}{4}}} \cdot {a^{\frac{7}{{12}}}}\left( {1 - a} \right)}} = \frac{{{a^{\frac{5}{6}}}\left( {1 + a} \right)}}{{{a^{\frac{5}{6}}}}} = 1 + a\)

Đó là các mặt phẳng ( SAC),(SBD) ,(SHJ), (SGI)  với G,H I,J , là các trung điểm của các cạnh AB , BC, CD,AD (hình vẽ bên dưới).

TXĐ: D = R.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaamyBaiabgkHiTiaabogacaqGVbGaae4CaiaadIhaaaa!3DAE! y' = m - {\rm{cos}}x\)

Hàm số đồng biến trên R

 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTabm % yEayaafaGaeyyzImRaaGimaiaacYcacqGHaiIicaWG4bGaeyicI4Sa % eSyhHekaaa!414B! \Leftrightarrow y' \ge 0,\forall x \in R\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % yBaiabgwMiZkGacohacaGGPbGaaiOBaiaadIhacaGGSaGaeyiaIiIa % amiEaiabgIGiolabl2riHcaa!444E! \Leftrightarrow m \ge \sin x,\forall x \in R\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % yBaiabgwMiZkaaigdaaaa!3BC3! \Leftrightarrow m \ge 1\)

TXĐ : D = R 

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsisl % caaI2aGaamiEaiabgkHiTiaaiMdaaaa!3F0B! y' = 3{x^2} - 6x - 9\) . Cho \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGimaiabgsDiBpaadeaaeaqabeaacaWG4bGaeyypa0Ja % eyOeI0IaaGymaaqaaiaadIhacqGH9aqpcaaIZaaaaiaawUfaaaaa!4287! y' = 0 \Leftrightarrow \left[ \begin{array}{l} x = - 1\\ x = 3 \end{array} \right.\)

Dựa vào đồ thị ta thấy hàm số đạt cực đại tại x = 0 và cực tiểu tại x = 2 

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGinaiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsisl % caaIXaGaaGOnaiaadIhaaaa!3E17! y' = 4{x^3} - 16x\), cho \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGimaiabgkDiElaaisdacaWG4bWaaWbaaSqabeaacaaI % ZaaaaOGaeyOeI0IaaGymaiaaiAdacaWG4bGaeyypa0JaaGimaiabgs % DiBpaadeaaeaqabeaacaWG4bGaeyypa0JaeyOeI0IaaGOmaiabgMGi % ppaadmaabaGaeyOeI0IaaGymaiaacUdacaaIXaaacaGLBbGaayzxaa % aabaGaamiEaiabg2da9iaaikdacqGHjiYZdaWadaqaaiabgkHiTiaa % igdacaGG7aGaaGymaaGaay5waiaaw2faaaqaaiaadIhacqGH9aqpca % aIWaGaeyicI48aamWaaeaacqGHsislcaaIXaGaai4oaiaaigdaaiaa % wUfacaGLDbaaaaGaay5waaaaaa!6341! y' = 0 \Rightarrow 4{x^3} - 16x = 0 \Leftrightarrow \left[ \begin{array}{l} x = - 2 \notin \left[ { - 1;1} \right]\\ x = 2 \notin \left[ { - 1;1} \right]\\ x = 0 \in \left[ { - 1;1} \right] \end{array} \right.\)

Khi đó: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaeyOeI0IaaGymaaGaayjkaiaawMcaaiabg2da9iaaigdacaaI % Waaaaa!3C8A! f\left( { - 1} \right) = 10\), \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaaGymaaGaayjkaiaawMcaaiabg2da9iaaigdacaaIWaaaaa!3B9D! f\left( 1 \right) = 10\),\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaaGimaaGaayjkaiaawMcaaiabg2da9iaaigdacaaI3aaaaa!3BA3! f\left( 0 \right) = 17\) .

Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGTbGaaiyyaiaacIhaaSqaamaadmaabaGaeyOeI0IaaGymaiaacUda % caaIXaaacaGLBbGaayzxaaaabeaakiaadMhacqGH9aqpcaWGMbWaae % WaaeaacaaIWaaacaGLOaGaayzkaaGaeyypa0JaaGymaiaaiEdaaaa!45D2! \mathop {\max }\limits_{\left[ { - 1;1} \right]} y = f\left( 0 \right) = 17\).

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaCa % aaleqabaGaaG4maaaakiabgkHiTiaaiodacaWG4bGaey4kaSIaaGOm % aiaad2gacqGH9aqpcaaIWaaaaa!3EDB! {x^3} - 3x + 2m = 0\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaey % OeI0IaamiEamaaCaaaleqabaGaaG4maaaakiabgUcaRiaaiodacaWG % 4bGaeyypa0JaaGOmaiaad2gacaaMc8oaaa!4208! \Leftrightarrow - {x^3} + 3x = 2m\,(*)\)

Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iabgkHiTiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaaI % ZaGaamiEaaaa!3D71! y = - {x^3} + 3x\) có đồ thị  (C) và đường thẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaacQ % dacaWG5bGaeyypa0JaaGOmaiaad2gaaaa!3B4C! d:y = 2m\)

Số nghiệm của phương trình (*) phụ thuộc vào số giao điểm của đồ thị hàm số (C) và đường thẳng d

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaeyOeI0IaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGc % cqGHRaWkcaaIZaaaaa!3D3C! y' = - 3{x^2} + 3\), cho \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGimaiabgsDiBlabgkHiTiaaiodacaWG4bWaaWbaaSqa % beaacaaIYaaaaOGaey4kaSIaaG4maiabg2da9iaaicdacqGHuhY2da % WabaabaeqabaGaamiEaiabg2da9iabgkHiTiaaigdaaeaacaWG4bGa % eyypa0JaaGymaaaacaGLBbaaaaa!4BD9! y' = 0 \Leftrightarrow - 3{x^2} + 3 = 0 \Leftrightarrow \left[ \begin{array}{l} x = - 1\\ x = 1 \end{array} \right.\)

Nhìn bảng biến thiên suy ra:

Phương trình (*) có ba nghiệm phân biệt khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % OmaiabgYda8iaaikdacaWGTbGaeyipaWJaaGOmaaaa!3C0E! - 2 < 2m < 2\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaey % OeI0IaaGymaiabgYda8iaad2gacqGH8aapcaaIXaaaaa!3DAC! \Leftrightarrow - 1 < m < 1\)

Ta có : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qamaaDa % aaleaacaWGUbaabaGaam4AaaaakiaadggadaahaaWcbeqaaiaad6ga % cqGHsislcaWGRbaaaOGaamOyamaaCaaaleqabaGaam4Aaaaakiabg2 % da9iaadoeadaqhaaWcbaGaaGOmaiaaigdaaeaacaWGRbaaaOGaamiE % amaaCaaaleqabaGaaGOmaiaaigdacqGHsislcaWGRbaaaOGaaiOlam % aabmaabaGaeyOeI0YaaSaaaeaacaaIYaaabaGaamiEamaaCaaaleqa % baGaaGOmaaaaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaadUgaaa % GccqGH9aqpdaqadaqaaiabgkHiTiaaikdaaiaawIcacaGLPaaadaah % aaWcbeqaaiaadUgaaaGccaWGdbWaa0baaSqaaiaaikdacaaIXaaaba % Gaam4AaaaakiaadIhadaahaaWcbeqaaiaaikdacaaIXaGaeyOeI0Ia % aG4maiaadUgaaaaaaa!5CCF!\ C_n^k{a^{n - k}}{b^k} = C_{21}^k{x^{21 - k}}.{\left( { - \frac{2}{{{x^2}}}} \right)^k} = {\left( { - 2} \right)^k}C_{21}^k{x^{21 - 3k}}\)

Theo yêu cầu bài toán \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaaG % OmaiaaigdacqGHsislcaaIZaGaam4Aaiabg2da9iaaicdacqGHuhY2 % caWGRbGaeyypa0JaaG4naaaa!4333! \Leftrightarrow 21 - 3k = 0 \Leftrightarrow k = 7\). Vậy hệ số cần tìm là .\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % OmamaaCaaaleqabaGaaG4naaaakiaadoeadaqhaaWcbaGaaGOmaiaa % igdaaeaacaaI3aaaaaaa!3BC1! - {2^7}C_{21}^7\)

Trường hợp m = -1 , suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaaikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGym % aaaa!3C40! y = 2{x^2} + 1\) Hàm số có điểm cực tiểu mà không có điểm cực đại nên loại m = -1 .

Trường hợp \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabgc % Mi5kabgkHiTiaaigdaaaa!3A54! m \ne - 1\)

Ta có:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGinamaabmaabaGaamyBaiabgUcaRiaaigdaaiaawIca % caGLPaaacaWG4bWaaWbaaSqabeaacaaIZaaaaOGaeyOeI0IaaGOmam % aabmaabaGaamyBaiabgkHiTiaaigdaaiaawIcacaGLPaaacaWG4baa % aa!4593! y' = 4\left( {m + 1} \right){x^3} - 2\left( {m - 1} \right)x\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0JaaG % OmaiaadIhadaWadaqaaiaaikdadaqadaqaaiaad2gacqGHRaWkcaaI % XaaacaGLOaGaayzkaaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgk % HiTmaabmaabaGaamyBaiabgkHiTiaaigdaaiaawIcacaGLPaaaaiaa % wUfacaGLDbaaaaa!4678! = 2x\left[ {2\left( {m + 1} \right){x^2} - \left( {m - 1} \right)} \right]\)

Xét \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGimaiabgsDiBpaadeaaeaqabeaacaWG4bGaeyypa0Ja % aGimaaqaaiaadEgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9a % qpcaaIYaWaaeWaaeaacaWGTbGaey4kaSIaaGymaaGaayjkaiaawMca % aiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsisldaqadaqaaiaad2 % gacqGHsislcaaIXaaacaGLOaGaayzkaaGaeyypa0JaaGimamaabmaa % baGaaiOkaaGaayjkaiaawMcaaaaacaGLBbaaaaa!531B! y' = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ g\left( x \right) = 2\left( {m + 1} \right){x^2} - \left( {m - 1} \right) = 0\left( * \right) \end{array} \right.\)

Vì hàm trùng phương luôn đạt cực trị tại điểm x = 0 nên để hàm số có một điểm cực đại mà không có điểm cực tiểu thì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaad2gacqGHRaWkcaaIXaGaeyipaWJaaGimaaqaaiabgkHiTiaa % d2gacqGHRaWkcaaIXaGaeyizImQaaGimaaaacaGL7baacqGHuhY2da % GabaabaeqabaGaamyBaiabgYda8iabgkHiTiaaigdaaeaacaWGTbae % aaaaaaaaa8qacqGHLjYSpaGaaGymaaaacaGL7baaaaa!4C09! \left\{ \begin{array}{l} m + 1 < 0\\ - m + 1 \le 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m < - 1\\ m \ge 1 \end{array} \right.\), suy ra không tồn tại m thỏa yêu cầu bài toán.

Điều kiện: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgc % Mi5kaaikdaaaa!3973! x \ne 2\)

Phương trình hoành độ giao điểm : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG4bGaey4kaSIaaGymaaqaaiaadIhacqGHsislcaaIYaaaaiabg2da % 9iabgkHiTiaaikdacaWG4bGaey4kaSIaamyBaiabgsDiBlaaikdaca % WG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0YaaeWaaeaacaWGTbGa % ey4kaSIaaG4maaGaayjkaiaawMcaaiaadIhacqGHRaWkcaaIYaGaam % yBaiabgUcaRiaaigdacqGH9aqpcaaIWaWaaeWaaeaacaGGQaaacaGL % OaGaayzkaaaaaa!53F3! \frac{{x + 1}}{{x - 2}} = - 2x + m \Leftrightarrow 2{x^2} - \left( {m + 3} \right)x + 2m + 1 = 0\left( * \right)\)

Theo yêu cầu bài toán \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aae % WaaeaacaGGQaaacaGLOaGaayzkaaaaaa!3A86! \Leftrightarrow \left( * \right)\) có hai nghiệm phân biệt khác 2.

 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaamaabmaabaGaamyBaiabgUcaRiaaiodaaiaawIcacaGL % PaaadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI0aGaaiOlaiaaik % dadaqadaqaaiaaikdacaWGTbGaey4kaSIaaGymaaGaayjkaiaawMca % aiabg6da+iaaicdaaeaacaaI4aGaeyOeI0IaaGOmamaabmaabaGaam % yBaiabgUcaRiaaiodaaiaawIcacaGLPaaacqGHRaWkcaaIYaGaamyB % aiabgUcaRiaaigdacqGHGjsUcaaIWaaaaiaawUhaaaaa!555F! \Leftrightarrow \left\{ \begin{array}{l} {\left( {m + 3} \right)^2} - 4.2\left( {2m + 1} \right) > 0\\ 8 - 2\left( {m + 3} \right) + 2m + 1 \ne 0 \end{array} \right.\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaad2gadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI % XaGaaGimaiaad2gacqGHRaWkcaaIXaGaeyOpa4JaaGimaaqaaiaaio % dacqGHGjsUcaaIWaaaaiaawUhaaiabgsDiBlaad2gacqGH8aapcaaI % 1aGaeyOeI0IaaGOmamaakaaabaGaaGOnaaWcbeaaaaa!4CDB! \Leftrightarrow \left\{ \begin{array}{l} {m^2} - 10m + 1 > 0\\ 3 \ne 0 \end{array} \right. \Leftrightarrow m < 5 - 2\sqrt 6 \ \)hoặc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabg6 % da+iaaiwdacqGHRaWkcaaIYaWaaOaaaeaacaaI2aaaleqaaaaa!3B25! m > 5 + 2\sqrt 6 \).

Xét phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG4bWaaWbaaSqabeaacaaIZaaaaOGaeyOeI0IaaG4maiaadIhadaah % aaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaaacaGLOaGaayzkaaWaaW % baaSqabeaacaaIZaaaaOGaeyOeI0IaaG4mamaabmaabaGaamiEamaa % CaaaleqabaGaaG4maaaakiabgkHiTiaaiodacaWG4bWaaWbaaSqabe % aacaaIYaaaaOGaey4kaSIaaGOmaaGaayjkaiaawMcaamaaCaaaleqa % baGaaGOmaaaakiabgUcaRiaaikdacqGH9aqpcaaIWaaaaa!4E47! {\left( {{x^3} - 3{x^2} + 2} \right)^3} - 3{\left( {{x^3} - 3{x^2} + 2} \right)^2} + 2 = 0\) (1)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIZaGaamiE % amaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaikdaaaa!3F1C! t = {x^3} - 3{x^2} + 2\) (*) thì  trở thành \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDamaaCa % aaleqabaGaaG4maaaakiabgkHiTiaaiodacaWG0bWaaWbaaSqabeaa % caaIYaaaaOGaey4kaSIaaGOmaiabg2da9iaaicdaaaa!3ED5! {t^3} - 3{t^2} + 2 = 0\) (2)

Theo đồ thị ta có (2) có ba nghiệm phân biệt  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamqaaqaabe % qaaiaadshacqGH9aqpcaaIXaaabaGaamiDaiabg2da9iaaigdacqGH % sisldaGcaaqaaiaaiodaaSqabaaakeaacaWG0bGaeyypa0JaaGymai % abgUcaRmaakaaabaGaaG4maaWcbeaaaaGccaGLBbaaaaa!42B8! \left[ \begin{array}{l} t = 1\\ t = 1 - \sqrt 3 \\ t = 1 + \sqrt 3 \end{array} \right.\)

Từ đồ thị hàm số ta có

+\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaaigdacqGHiiIZdaqadaqaaiabgkHiTiaaikdacaGG7aGaaGOm % aaGaayjkaiaawMcaaaaa!3EDF! t = 1 \in \left( { - 2;2} \right)\)(*) có ba nghiệm phân biệt

+ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaaigdacqGHsisldaGcaaqaaiaaiodaaSqabaGccqGHiiIZdaqa % daqaaiabgkHiTiaaikdacaGG7aGaaGOmaaGaayjkaiaawMcaaaaa!40AE! t = 1 - \sqrt 3 \in \left( { - 2;2} \right)\) nên (*) có ba nghiệm phân biệt (khác ba nghiệm khi t = 1)

+ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaaigdacqGHRaWkdaGcaaqaaiaaiodaaSqabaGccqGH+aGpcaaI % Yaaaaa!3C36! t = 1 + \sqrt 3 > 2\) nên (*) có đúng một nghiệm

Vậy phương trình đã cho có 7 nghiệm phân biệt

Nhận xét: Với mỗi giá trị t , học sinh có thể sử dụng máy tính bỏ túi để thử nghiệm

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaad2gadaqadaqaaiaa % dIhacqGHsislcaaIXaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYa % aaaOGaey4kaSIaaGinaiabg2da9iaad2gacaWG4bWaaWbaaSqabeaa % caaIYaaaaOGaeyOeI0IaaGOmaiaad2gacaWG4bGaey4kaSIaaGinai % abgUcaRiaad2gaaaa!4D13! g\left( x \right) = m{\left( {x - 1} \right)^2} + 4 = m{x^2} - 2mx + 4 + m\)

Để đồ thị hàm số có hai tiệm cận đứng thì cần tìm m để phương trình g(x) = 0 có hai nghiệm phân biệt khác  - 1

ĐK: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaad2gacqGHGjsUcaaIWaaabaGaeuiLdqKaeyypa0JaamyBamaa % CaaaleqabaGaaGOmaaaakiabgkHiTiaad2gadaqadaqaaiaaisdacq % GHRaWkcaWGTbaacaGLOaGaayzkaaGaeyOpa4JaaGimaaqaaiaadEga % daqadaqaaiabgkHiTiaaigdaaiaawIcacaGLPaaacqGHGjsUcaaIWa % aaaiaawUhaaaaa!4D34! \left\{ \begin{array}{l} m \ne 0\\ \Delta = {m^2} - m\left( {4 + m} \right) > 0\\ g\left( { - 1} \right) \ne 0 \end{array} \right.\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaad2gacqGH8aapcaaIWaaabaGaamyBaiabgcMi5kab % gkHiTiaaigdaaaGaay5Eaaaaaa!4082! \Leftrightarrow \left\{ \begin{array}{l} m < 0\\ m \ne - 1 \end{array} \right.\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iabgkHiTmaabmaabaGaamiEamaaCaaaleqabaGaaGOmaaaakiab % gkHiTiaaigdaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccq % GHsislcaaIXaGaeyizImQaeyOeI0IaaGymaiaacYcacqGHaiIicaWG % 4bGaeyicI4SaeSyhHekaaa!496F! y = - {\left( {{x^2} - 1} \right)^2} - 1 \le - 1,\forall x \in R\). Do đó đồ thị của hàm số này nằm dưới Ox

Nhận xét có thể lập bảng biến thiên và kết luận.

Do đồ thị cắt Oy tại M(0;c) nằm dưới trục Ox nên c < 0

Vì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhaqaaaaaaaaaWdbiabgkziUkab % gglaXkabg6HiLcWdaeqaaOGaamyEaiabg2da9iabgUcaR8qacqGHEi % sPaaa!43E5! \mathop {\lim }\limits_{x \to \pm \infty } y = + \infty \) nên a > 0 .

Hàm số có ba điểm cực trị nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiaadk % gacqGH8aapcaaIWaGaeyO0H4TaamOyaiabgYda8iaaicdaaaa!3E81! ab < 0 \Rightarrow b < 0\)

Xét  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIZaGaamiE % amaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaikdacaGGUaaaaa!3FD3! y = {x^3} - 3{x^2} + 2.\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaG4maiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsisl % caaI2aGaamiEaaaa!3D5B! y' = 3{x^2} - 6x\); \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaaGimaiabgsDiBpaadeaaeaqabeaacaWG4bGaeyypa0Ja % aGimaaqaaiaadIhacqGH9aqpcaaIYaaaaiaawUfaaaaa!4198! y' = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 2 \end{array} \right.\) . Khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iaaicdacqGHshI3caWG5bGaeyypa0JaaGOmaiaacUdacaaMe8Ua % amiEaiabg2da9iaaikdacqGHshI3caWG5bGaeyypa0JaeyOeI0IaaG % Omaaaa!48E3! x = 0 \Rightarrow y = 2;\;x = 2 \Rightarrow y = - 2\)

Hàm số này thỏa mãn các tính chất trên bảng biến thiên.

Từ đồ thị thấy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaaGimaiabgsDi % BpaadeaaeaqabeaacaWG4bGaeyypa0JaeyOeI0IaaGymaaqaaiaadI % hacqGH9aqpcaaIYaaaaiaawUfaaaaa!44F9! f'\left( x \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = - 1\\ x = 2 \end{array} \right.\) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyOpa4JaaGimaiabgsDi % BlaadIhacqGH+aGpcaaIYaaaaa!4050! f'\left( x \right) > 0 \Leftrightarrow x > 2\)

Xét \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadAgadaqadaqaaiaa % dIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaIYaaacaGLOaGaay % zkaaaaaa!4079! g\left( x \right) = f\left( {{x^2} - 2} \right)\) có TXĐ : D = R

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4zayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaaGOmaiaadIha % ceWGMbGbauaadaqadaqaaiaadshaaiaawIcacaGLPaaaaaa!3FAA! g'\left( x \right) = 2xf'\left( t \right)\) với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaIYaaaaa!3B8C! t = {x^2} - 2\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4zayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaaGimaiabgsDi % BpaadeaaeaqabeaacaWG4bGaeyypa0JaaGimaaqaaiaadshacqGH9a % qpcaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOmaiabg2da % 9iabgkHiTiaaigdaaeaacaWG0bGaeyypa0JaamiEamaaCaaaleqaba % GaaGOmaaaakiabgkHiTiaaikdacqGH9aqpcaaIYaaaaiaawUfaaiab % gsDiBpaadeaaeaqabeaacaWG4bGaeyypa0JaaGimaaqaaiaadIhacq % GH9aqpcqGHXcqScaaIXaaabaGaamiEaiabg2da9iabgglaXkaaikda % aaGaay5waaaaaa!6063! g'\left( x \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ t = {x^2} - 2 = - 1\\ t = {x^2} - 2 = 2 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = \pm 1\\ x = \pm 2 \end{array} \right.\)

Có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG0baacaGLOaGaayzkaaGaeyOpa4JaaGimaiabgsDi % BlaadshacqGH9aqpcaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0 % IaaGOmaiabg6da+iaaikdacqGHuhY2caWG4bGaeyipaWJaeyOeI0Ia % aGOmaiaaykW7caaMc8UaeyikIOTaaGPaVlaaykW7caWG4bGaeyOpa4 % JaaGOmaaaa!558A! f'\left( t \right) > 0 \Leftrightarrow t = {x^2} - 2 > 2 \Leftrightarrow x < - 2\,\, \vee \,\,x > 2\)

Ta có : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaadggaaeqaaOGaamOyaiabg6da+iaaicda % cqGHuhY2daWabaabaeqabaWaaiqaaqaabeqaaiaadggacqGH+aGpca % aIXaaabaGaamOyaiabg6da+iaadggadaahaaWcbeqaaiaaicdaaaGc % cqGH9aqpcaaIXaaaaiaawUhaaaqaamaaceaaeaqabeaacaaIWaGaey % ipaWJaamyyaiabgYda8iaaigdaaeaacaaIWaGaeyipaWJaamOyaiab % gYda8iaadggadaahaaWcbeqaaiaaicdaaaGccqGH9aqpcaaIXaaaai % aawUhaaaaacaGLBbaaaaa!55FD! {\log _a}b > 0 \Leftrightarrow \left[ \begin{array}{l} \left\{ \begin{array}{l} a > 1\\ b > {a^0} = 1 \end{array} \right.\\ \left\{ \begin{array}{l} 0 < a < 1\\ 0 < b < {a^0} = 1 \end{array} \right. \end{array} \right.\)Vậy chọn B

Điều kiện: x > 0 .

PT: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaci % iBaiaac+gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaadaWc % aaqaaiaaikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaG % ymaaqaaiaaikdacaWG4baaaaGaayjkaiaawMcaaiabgUcaRiaaikda % daahaaWcbeqaamaabmaabaWaaSaaaeaacaaIYaGaamiEamaaCaaame % qabaGaaGOmaaaaliabgUcaRiaaigdaaeaacaaIYaGaamiEaaaaaiaa % wIcacaGLPaaaaaGccqGH9aqpcaaI1aGaaGPaVlaaykW7caaMc8UaaG % PaVlaaykW7daqadaqaaiaaigdaaiaawIcacaGLPaaaaaa!58DF! \Leftrightarrow {\log _2}\left( {\frac{{2{x^2} + 1}}{{2x}}} \right) + {2^{\left( {\frac{{2{x^2} + 1}}{{2x}}} \right)}} = 5\,\,\,\,\,\left( 1 \right)\)

Đặt  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9maalaaabaGaaGOmaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGH % RaWkcaaIXaaabaGaaGOmaiaadIhaaaGaeyypa0JaamiEaiabgUcaRm % aalaaabaGaaGymaaqaaiaaikdacaWG4baaaiabgwMiZkaaikdadaGc % aaqaaiaadIhacaGGUaWaaSaaaeaacaaIXaaabaGaaGOmaiaadIhaaa % aaleqaaOGaeyypa0ZaaOaaaeaacaaIYaaaleqaaaaa!4C25! t = \frac{{2{x^2} + 1}}{{2x}} = x + \frac{1}{{2x}} \ge 2\sqrt {x.\frac{1}{{2x}}} = \sqrt 2 \)

PT trở thành \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOGaamiDaiabgUcaRiaaikda % daahaaWcbeqaaiaadshaaaGccqGH9aqpcaaI1aGaaeiiaiaabccaca % qGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaa % bccacaqGGaGaaeiiaiaabccacaqGOaGaaeOmaiaabMcaaaa!4A38! {\log _2}t + {2^t} = 5{\rm { (2)}}\)

Xét hàm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiDaaGaayjkaiaawMcaaiabg2da9iGacYgacaGGVbGaai4z % amaaBaaaleaacaaIYaaabeaakiaadshacqGHRaWkcaaIYaWaaWbaaS % qabeaacaWG0baaaOGaaGPaVlaaykW7daqadaqaaiaadshacqGHLjYS % daGcaaqaaiaaikdaaSqabaaakiaawIcacaGLPaaaaaa!4A2F! f\left( t \right) = {\log _2}t + {2^t}\,\,\left( {t \ge \sqrt 2 } \right)\) là hàm đồng biến nên:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIYaaacaGLOaGaayzkaaGaeyi1HSTaamOzamaabmaabaGaamiDaaGa % ayjkaiaawMcaaiabg2da9iaadAgadaqadaqaaiaaikdaaiaawIcaca % GLPaaacqGHuhY2caWG0bGaeyypa0JaaGOmaaaa!474F! \left( 2 \right) \Leftrightarrow f\left( t \right) = f\left( 2 \right) \Leftrightarrow t = 2\) (t/m).

Với t = 2 thì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIYaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaigdaaeaa % caaIYaGaamiEaaaacqGH9aqpcaaIYaGaeyi1HSTaaGOmaiaadIhada % ahaaWcbeqaaiaaikdaaaGccqGHsislcaaI0aGaamiEaiabgUcaRiaa % igdacqGH9aqpcaaIWaaaaa!48D5! \frac{{2{x^2} + 1}}{{2x}} = 2 \Leftrightarrow 2{x^2} - 4x + 1 = 0\)  (t/m). Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBa % aaleaacaaIXaaabeaakiaadIhadaWgaaWcbaGaaGOmaaqabaGccqGH % 9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaaaaa!3C5E! {x_1}{x_2} = \frac{1}{2}\)  (theo Viet ).

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: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiraiabg2 % da9maabmaabaGaaGymaiaacUdacaaMc8Uaey4kaSIaeyOhIukacaGL % OaGaayzkaaaaaa!3EA4! D = \left( {1;\, + \infty } \right)\).

Xét hai khai triển:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmamaaCa % aaleqabaGaaGOmaiaaicdacaaIXaGaaG4naaaakiabg2da9maabmaa % baGaaGymaiabgUcaRiaaigdaaiaawIcacaGLPaaadaahaaWcbeqaai % aaikdacaaIWaGaaGymaiaaiEdaaaGccqGH9aqpcaWGdbWaa0baaSqa % aiaaikdacaaIWaGaaGymaiaaiEdaaeaacaaIWaaaaOGaey4kaSIaam % 4qamaaDaaaleaacaaIYaGaaGimaiaaigdacaaI3aaabaGaaGymaaaa % kiabgUcaRiaadoeadaqhaaWcbaGaaGOmaiaaicdacaaIXaGaaG4naa % qaaiaaikdaaaGccqGHRaWkcaWGdbWaa0baaSqaaiaaikdacaaIWaGa % aGymaiaaiEdaaeaacaaIZaaaaOGaey4kaSIaaiOlaiaac6cacaGGUa % Gaey4kaSIaam4qamaaDaaaleaacaaIYaGaaGimaiaaigdacaaI3aaa % baGaaGOmaiaaicdacaaIXaGaaG4naaaakiaaykW7caaMc8UaaGPaVp % aabmaabaGaaGymaaGaayjkaiaawMcaaaaa!69E9! {2^{2017}} = {\left( {1 + 1} \right)^{2017}} = C_{2017}^0 + C_{2017}^1 + C_{2017}^2 + C_{2017}^3 + ... + C_{2017}^{2017}\,\,\,\left( 1 \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGimaiabg2 % da9maabmaabaGaaGymaiabgkHiTiaaigdaaiaawIcacaGLPaaadaah % aaWcbeqaaiaaikdacaaIWaGaaGymaiaaiEdaaaGccqGH9aqpcaWGdb % Waa0baaSqaaiaaikdacaaIWaGaaGymaiaaiEdaaeaacaaIWaaaaOGa % eyOeI0Iaam4qamaaDaaaleaacaaIYaGaaGimaiaaigdacaaI3aaaba % GaaGymaaaakiabgUcaRiaadoeadaqhaaWcbaGaaGOmaiaaicdacaaI % XaGaaG4naaqaaiaaikdaaaGccqGHsislcaWGdbWaa0baaSqaaiaaik % dacaaIWaGaaGymaiaaiEdaaeaacaaIZaaaaOGaey4kaSIaaiOlaiaa % c6cacaGGUaGaeyOeI0Iaam4qamaaDaaaleaacaaIYaGaaGimaiaaig % dacaaI3aaabaGaaGOmaiaaicdacaaIXaGaaG4naaaakiaaykW7caaM % c8UaaGPaVlaaykW7daqadaqaaiaaikdaaiaawIcacaGLPaaaaaa!6876! 0 = {\left( {1 - 1} \right)^{2017}} = C_{2017}^0 - C_{2017}^1 + C_{2017}^2 - C_{2017}^3 + ... - C_{2017}^{2017}\,\,\,\,\left( 2 \right)\)

Lấy (1) - (2) theo vế ta được: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmamaaCa % aaleqabaGaaGOmaiaaicdacaaIXaGaaG4naaaakiabg2da9iaaykW7 % caaMc8UaaGPaVlaaikdadaqadaqaaiaadoeadaqhaaWcbaGaaGOmai % aaicdacaaIXaGaaG4naaqaaiaaigdaaaGccqGHRaWkcaWGdbWaa0ba % aSqaaiaaikdacaaIWaGaaGymaiaaiEdaaeaacaaIZaaaaOGaey4kaS % Iaam4qamaaDaaaleaacaaIYaGaaGimaiaaigdacaaI3aaabaGaaGyn % aaaakiabgUcaRiaac6cacaGGUaGaaiOlaiabgUcaRiaadoeadaqhaa % WcbaGaaGOmaiaaicdacaaIXaGaaG4naaqaaiaaikdacaaIWaGaaGym % aiaaiEdaaaaakiaawIcacaGLPaaacqGHshI3caWGubGaeyypa0JaaG % OmamaaCaaaleqabaGaaGOmaiaaicdacaaIXaGaaGOnaaaaaaa!6466! {2^{2017}} = \,\,\,2\left( {C_{2017}^1 + C_{2017}^3 + C_{2017}^5 + ... + C_{2017}^{2017}} \right) \Rightarrow T = {2^{2016}}\)

Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacYgacaGGVbGaai4zamaaBaaaleaadaWcaaqaaiaaigdaaeaa % caaIYaaaaaqabaGccaWG4baaaa!3D82! y = {\log _{\frac{1}{2}}}x\) có TXĐ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiraiabg2 % da9maabmaabaGaaGimaiaacUdacqGHRaWkcqGHEisPaiaawIcacaGL % Paaaaaa!3D18! D = \left( {0; + \infty } \right)\) nên không thỏa mãn.

Do \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaacq % aHapaCaeaacaaIZaaaaiabg6da+iaaigdaaaa!3A40! \frac{\pi }{3} > 1\) nên hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maabmaabaWaaSaaaeaacqaHapaCaeaacaaIZaaaaaGaayjkaiaa % wMcaamaaCaaaleqabaGaamiEaaaaaaa!3D35! y = {\left( {\frac{\pi }{3}} \right)^x}\) đồng biến trên R.

Do \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGimaiabgY % da8maalaaabaGaaGOmaaqaaiaadwgaaaGaeyipaWJaaGymaaaa!3B27! 0 < \frac{2}{e} < 1\) nên hàm số nghịch biến trên R

Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iGacYgacaGGVbGaai4zamaaBaaaleaadaWcaaqaaiabec8aWbqa % aiaaisdaaaaabeaakmaabmaabaGaaGOmaiaadIhadaahaaWcbeqaai % aaikdaaaGccqGHRaWkcaaIXaaacaGLOaGaayzkaaaaaa!435B! y = {\log _{\frac{\pi }{4}}}\left( {2{x^2} + 1} \right)\) có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0ZaaSaaaeaacaaI0aGaamiEaaqaamaabmaabaGaaGOmaiaa % dIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIXaaacaGLOaGaay % zkaaGaciiBaiaac6gadaqadaqaamaalaaabaGaeqiWdahabaGaaGin % aaaaaiaawIcacaGLPaaaaaaaaa!4598! y' = \frac{{4x}}{{\left( {2{x^2} + 1} \right)\ln \left( {\frac{\pi }{4}} \right)}}\) đổi dấu khi x đi qua 0 nên không nghịch biến trên R .

Gọi O;O' là tâm của hai đáy của hình trụ và (P) là mặt phẳng song song với trục và cách trục một khoảng 3cm.

Mp(P) cắt hai hình tròn đáy (O);(O') theo hai dây cung lần lượt là AB và CD cắt mặt xung quanh theo hai đường sinh là AD,BC. Khi đó ABCD là hình chữ nhật.

Gọi H là trung điểm của AB. Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4taiaadI % eacqGHLkIxcaWGbbGaamOqaiaacUdacaWGpbGaamisaiabgwQiEjaa % dgeacaWGebGaeyO0H4Taam4taiaadIeacqGHLkIxdaqadaqaaiaadg % eacaWGcbGaam4qaiaadseaaiaawIcacaGLPaaaaaa!4AC9! OH \bot AB;OH \bot AD \Rightarrow OH \bot \left( {ABCD} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % izamaabmaabaGaam4taiaaykW7ceWGpbGbauaacaGGSaWaaeWaaeaa % caWGqbaacaGLOaGaayzkaaaacaGLOaGaayzkaaGaeyypa0Jaamizam % aabmaabaGaam4taiaacYcadaqadaqaaiaadgeacaWGcbGaam4qaiaa % dseaaiaawIcacaGLPaaaaiaawIcacaGLPaaacqGH9aqpcaWGpbGaam % isaiabg2da9iaaiodacaqGJbGaaeyBaaaa!50F3! \Rightarrow d\left( {O\,O',\left( P \right)} \right) = d\left( {O,\left( {ABCD} \right)} \right) = OH = 3{\rm{cm}}\)

Khi đó:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadk % eacqGH9aqpcaaIYaGaamyqaiaadIeacqGH9aqpcaaIYaWaaOaaaeaa % caWGpbGaamyqamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaad+eaca % WGibWaaWbaaSqabeaacaaIYaaaaaqabaGccqGH9aqpcaaIYaWaaOaa % aeaacaaI1aWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaG4mamaaCa % aaleqabaGaaGOmaaaaaeqaaOGaeyypa0JaaGioaaaa!4A9F! AB = 2AH = 2\sqrt {O{A^2} - O{H^2}} = 2\sqrt {{5^2} - {3^2}} = 8\);\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaads % eacqGH9aqpcaWGpbGaaGPaVlaad+eacaGGNaGaeyypa0JaamiAaiab % g2da9iaaiEdacaqGJbGaaeyBaaaa!41F7! AD = O\,O' = h = 7{\rm{cm}}\)

Diện tích hình chữ nhật ABCD là: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaWGbbGaamOqaiaadoeacaWGebaabeaakiabg2da9iaadgea % caWGcbGaaiOlaiaadgeacaWGebGaeyypa0JaaGynaiaaiAdadaqada % qaaiaadogacaWGTbWaaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzk % aaaaaa!45CF! {S_{ABCD}} = AB.AD = 56\left( {c{m^2}} \right)\)

Đường cao lăng trụ là AD = AB= 30cm không đổi. Để thể tích lăng trụ lớn nhất chỉ cần diện tích đáy lớn nhất.

Gọi I là trung điểm cạnh EG \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yqaiaadMeacqGHLkIxcaWGfbGaam4raaaa!3D2C! \Rightarrow AI \bot EG\)  trong tam giác AEG

Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiaadE % eacqGH9aqpcaaIXaGaaGynaiabgkHiTiaadIhacaGGSaaaaa!3CA8! IG = 15 - x,\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIWaGaeyipaWJaamiEaiabgYda8iaaigdacaaI1aaacaGLOaGaayzk % aaaaaa!3CB6! \left( {0 < x < 15} \right)\)

Có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadM % eacqGH9aqpdaGcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGH % sisldaqadaqaamaalaaabaGaaG4maiaaicdacqGHsislcaaIYaGaam % iEaaqaaiaaikdaaaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaa % aaqabaGccqGH9aqpdaGcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaa % GccqGHsisldaqadaqaaiaaigdacaaI1aGaeyOeI0IaamiEaaGaayjk % aiaawMcaamaaCaaaleqabaGaaGOmaaaaaeqaaaaa!4CA8! AI = \sqrt {{x^2} - {{\left( {\frac{{30 - 2x}}{2}} \right)}^2}} = \sqrt {{x^2} - {{\left( {15 - x} \right)}^2}} \)   \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaO % aaaeaacaaIZaGaaGimaiaadIhacqGHsislcaaIYaGaaGOmaiaaiwda % aSqabaGccaGGSaGaaGPaVlaadIhacqGHiiIZdaqadaqaamaalaaaba % GaaGymaiaaiwdaaeaacaaIYaaaaiaacUdacaaIXaGaaGynaaGaayjk % aiaawMcaaaaa!477A! = \sqrt {30x - 225} ,\,x \in \left( {\frac{{15}}{2};15} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacqqHuoarcaWGbbGaamyraiaadEeaaeqaaOGaeyypa0ZaaSaa % aeaacaaIXaaabaGaaGOmaaaacaWGbbGaamysaiaac6cacaWGfbGaam % 4raiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaWaaeWaaeaacaaI % ZaGaaGimaiabgkHiTiaaikdacaWG4baacaGLOaGaayzkaaWaaOaaae % aacaaIZaGaaGimaiaadIhacqGHsislcaaIYaGaaGOmaiaaiwdaaSqa % baaaaa!4F12! {S_{\Delta AEG}} = \frac{1}{2}AI.EG = \frac{1}{2}\left( {30 - 2x} \right)\sqrt {30x - 225} \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaO % aaaeaacaaIXaGaaGynaaWcbeaakiaac6cadaGcaaqaamaabmaabaGa % aGymaiaaiwdacqGHsislcaWG4baacaGLOaGaayzkaaWaaWbaaSqabe % aacaaIYaaaaOWaaeWaaeaacaaIYaGaamiEaiabgkHiTiaaigdacaaI % 1aaacaGLOaGaayzkaaaaleqaaaaa!44EF! = \sqrt {15} .\sqrt {{{\left( {15 - x} \right)}^2}\left( {2x - 15} \right)}\)

Vậy ta cần tìm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaabmaabaWaaSaaaeaacaaIXaGaaGynaaqaaiaaikdaaaGaai4o % aiaaigdacaaI1aaacaGLOaGaayzkaaaaaa!3E7D! x \in \left( {\frac{{15}}{2};15} \right)\) để \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maabmaabaGaaGymaiaa % iwdacqGHsislcaWG4baacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYa % aaaOWaaeWaaeaacaaIYaGaamiEaiabgkHiTiaaigdacaaI1aaacaGL % OaGaayzkaaaaaa!45F4! f\left( x \right) = {\left( {15 - x} \right)^2}\left( {2x - 15} \right)\) lớn nhất.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaeyOeI0IaaGOm % amaabmaabaGaaGymaiaaiwdacqGHsislcaWG4baacaGLOaGaayzkaa % WaaeWaaeaacaaIYaGaamiEaiabgkHiTiaaigdacaaI1aaacaGLOaGa % ayzkaaGaey4kaSIaaGOmamaabmaabaGaaGymaiaaiwdacqGHsislca % WG4baacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaeyypa0Ja % aGOmamaabmaabaGaaGymaiaaiwdacqGHsislcaWG4baacaGLOaGaay % zkaaWaaeWaaeaacaaIZaGaaGimaiabgkHiTiaaiodacaWG4baacaGL % OaGaayzkaaGaeyypa0JaaGimaiabgsDiBpaadeaaeaqabeaacaWG4b % Gaeyypa0JaaGymaiaaiwdaaeaacaWG4bGaeyypa0JaaGymaiaaicda % aaGaay5waaaaaa!669D! f'\left( x \right) = - 2\left( {15 - x} \right)\left( {2x - 15} \right) + 2{\left( {15 - x} \right)^2} = 2\left( {15 - x} \right)\left( {30 - 3x} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 15\\ x = 10 \end{array} \right.\)

Vậy thể tích lăng trụ lớn nhất khi x = 10

Cách khác (trắc nghiệm): Học sinh có thể thay giá trị của từng đáp án vào hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maabmaabaGaaGymaiaa % iwdacqGHsislcaWG4baacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYa % aaaOWaaeWaaeaacaaIYaGaamiEaiabgkHiTiaaigdacaaI1aaacaGL % OaGaayzkaaaaaa!45F4! f\left( x \right) = {\left( {15 - x} \right)^2}\left( {2x - 15} \right)\) để có kết quả.

Đk:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaadmaabaGaaGimaiaacUdacaaIXaGaaGynaaGaay5waiaaw2fa % aaaa!3D59! x \in \left[ {0;15} \right]\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaadmaabaGaaGimaiaacUdacaaIXaGaaGynaaGaay5waiaaw2fa % aaaa!3D59! x \in \left[ {0;15} \right]\). (vì độ giảm huyết áp không thể là số âm)

Có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4rayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaaGimaiaacYca % caaIWaGaaG4maiaaiwdadaWadaqaaiaaikdacaWG4bWaaeWaaeaaca % aIXaGaaGynaiabgkHiTiaadIhaaiaawIcacaGLPaaacqGHsislcaWG % 4bWaaWbaaSqabeaacaaIYaaaaaGccaGLBbGaayzxaaGaeyypa0JaaG % imaiaacYcacaaIXaGaaGimaiaaiwdacaWG4bWaaeWaaeaacaaIXaGa % aGimaiabgkHiTiaadIhaaiaawIcacaGLPaaacqGH9aqpcaaIWaGaey % i1HS9aamqaaqaabeqaaiaadIhacqGH9aqpcaaIWaaabaGaamiEaiab % g2da9iaaigdacaaIWaaaaiaawUfaaaaa!5F48! G'\left( x \right) = 0,035\left[ {2x\left( {15 - x} \right) - {x^2}} \right] = 0,105x\left( {10 - x} \right) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0\\ x = 10 \end{array} \right.\)

\(G(0) = 0 \)  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4ramaabm % aabaGaaGymaiaaicdaaiaawIcacaGLPaaacqGH9aqpdaWcaaqaaiaa % iodacaaI1aaabaGaaGOmaaaaaaa!3D0C! ; G\left( {10} \right) = \frac{{35}}{2}\)\(; G(15) = 0\)

Vậy huyết áp bệnh nhân giảm nhiều nhất khi tiêm cho bệnh nhân liều

x = 10 miligam.

Có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaiwdaaeqaaOGaaGinaiabg2da9iaadkga % aaa!3C64! {\log _5}4 = b\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aaS % aaaeaacaaIYaGaciiBaiaac6gacaaIYaaabaGaciiBaiaac6gacaaI % 1aaaaiabg2da9iaadkgacqGHuhY2ciGGSbGaaiOBaiaaiwdacqGH9a % qpdaWcaaqaaiaaikdacaWGHbaabaGaamOyaaaaaaa!48EA! \Leftrightarrow \frac{{2\ln 2}}{{\ln 5}} = b \Leftrightarrow \ln 5 = \frac{{2a}}{b}\)

Khi đó: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6 % gacaaIXaGaaGimaiaaicdacqGH9aqpcaaIYaGaciiBaiaac6gacaaI % XaGaaGimaiabg2da9iaaikdadaqadaqaaiGacYgacaGGUbGaaGOmai % abgUcaRiGacYgacaGGUbGaaGynaaGaayjkaiaawMcaaaaa!4892! \ln 100 = 2\ln 10 = 2\left( {\ln 2 + \ln 5} \right)\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0JaaG % OmamaabmaabaGaamyyaiabgUcaRmaalaaabaGaaGOmaiaadggaaeaa % caWGIbaaaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaGOmaiaadg % gacaWGIbGaey4kaSIaaGinaiaadggaaeaacaWGIbaaaaaa!44AC! = 2\left( {a + \frac{{2a}}{b}} \right) = \frac{{2ab + 4a}}{b}\)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaaikdadaahaaWcbeqaaiaadIhaaaGccaGGSaGaaGPaVlaadsha % cqGH+aGpcaaIWaaaaa!3ED8! t = {2^x},\,t > 0\) 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 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmamaaCa % aaleqabaGaamiEaaaakiabg2da9iaaiodacqGHuhY2caWG4bGaeyyp % a0JaciiBaiaac+gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOGaaG4maa % aa!4284! {2^x} = 3 \Leftrightarrow x = {\log _2}3\)

Để được một số có 4 chữ số theo yêu cầu đề bài, ta chọn 4 chữ số trong 6 chữ số đã cho và xếp theo một thứ tự nào đó, nghĩa là ta được một chỉnh hợp chập 4 của 6 phần tử.

Vậy số các số cần thành lập là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqamaaDa % aaleaacaaI2aaabaGaaGinaaaakiabg2da9iaaiodacaaI2aGaaGim % aaaa!3BAB! A_6^4 = 360\)

Gọi O là tâm của \(\Delta ABC\) suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaad+ % eacqGHLkIxdaqadaqaaiaadgeacaWGcbGaam4qaaGaayjkaiaawMca % aaaa!3D2E! SO \bot \left( {ABC} \right)\) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaad+ % eacqGH9aqpcaWGObGaeyypa0JaaGymaaaa!3B53! SO = h = 1\); \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4taiaadg % eacqGH9aqpdaWcaaqaaiaaikdaaeaacaaIZaaaaiabgwSixpaakaaa % baGaaGOnaaWcbeaakiabgwSixpaalaaabaWaaOaaaeaacaaIZaaale % qaaaGcbaGaaGOmaaaacqGH9aqpdaGcaaqaaiaaikdaaSqabaaaaa!4320! OA = \frac{2}{3} \cdot \sqrt 6 \cdot \frac{{\sqrt 3 }}{2} = \sqrt 2 \)

Trong tam giác vuông SAO, ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaadg % eacqGH9aqpdaGcaaqaaiaadofacaWGpbWaaWbaaSqabeaacaaIYaaa % aOGaey4kaSIaam4taiaadgeadaahaaWcbeqaaiaaikdaaaaabeaaki % abg2da9maakaaabaGaaGymaiabgUcaRiaaikdaaSqabaGccqGH9aqp % daGcaaqaaiaaiodaaSqabaaaaa!4417! SA = \sqrt {S{O^2} + O{A^2}} = \sqrt {1 + 2} = \sqrt 3 \)

Trong mặt phẳng (SAO) kẻ trung trực của đoạn SA cắt SO tại I , suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiaado % facqGH9aqpcaWGjbGaamyqaiabg2da9iaadMeacaWGcbGaeyypa0Ja % amysaiaadoeaaaa!3F6A! IS = IA = IB = IC\) nên  là tâm mặt cầu ngoại tiếp hình chóp .

Gọi H là trung điểm của SA, ta có \(\Delta SHI\) đồng dạng với \(\Delta SOA\) nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabg2 % da9iaadMeacaWGtbGaeyypa0ZaaSaaaeaacaWGtbGaamisaiaac6ca % caWGtbGaamyqaaqaaiaadofacaWGpbaaaiabg2da9maalaaabaWaaS % aaaeaadaGcaaqaaiaaiodaaSqabaaakeaacaaIYaaaaiabgwSixpaa % kaaabaGaaG4maaWcbeaaaOqaaiaaigdaaaGaeyypa0ZaaSaaaeaaca % aIZaaabaGaaGOmaaaaaaa!4967! R = IS = \frac{{SH.SA}}{{SO}} = \frac{{\frac{{\sqrt 3 }}{2} \cdot \sqrt 3 }}{1} = \frac{3}{2}\)

. Vậy diện tích mặt cầu \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaWGTbGaam4yaaqabaGccqGH9aqpcaaI0aGaeqiWdaNaamOu % amaaCaaaleqabaGaaGOmaaaakiabg2da9iaaiMdacqaHapaCaaa!41AC! {S_{mc}} = 4\pi {R^2} = 9\pi \)

Số hạng tổng quát trong khai triển nhị thức Newton \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIYaGaeyOeI0IaamiEaaGaayjkaiaawMcaamaaCaaaleqabaGaamOB % aaaakiaacYcacaaMc8+aaeWaaeaacaWGUbGaeyicI4SaeSyfHu6aaW % baaSqabeaacaGGQaaaaaGccaGLOaGaayzkaaaaaa!43D8! {\left( {2 - x} \right)^n},\,\left( {n \in {N^*}} \right)\) là

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qamaaDa % aaleaacaWGUbaabaGaam4AaaaakiaaikdadaahaaWcbeqaaiaad6ga % cqGHsislcaWGRbaaaOWaaeWaaeaacqGHsislcaaIXaaacaGLOaGaay % zkaaWaaWbaaSqabeaacaWGRbaaaOGaamiEamaaCaaaleqabaGaam4A % aaaaaaa!430A! C_n^k{2^{n - k}}{\left( { - 1} \right)^k}{x^k}\), với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4AaiabgI % GiolablssiIkaacYcacaaMc8UaaGPaVlaaicdacqGHKjYOcaWGRbGa % eyizImQaamOBaaaa!43AC! k \in Z ,\,\,0 \le k \le n\), suy ra hệ số của \(x^4\) là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qamaaDa % aaleaacaWGUbaabaGaaGinaaaakiaaikdadaahaaWcbeqaaiaad6ga % cqGHsislcaaI0aaaaaaa!3C2A! C_n^4{2^{n - 4}}\). Theo đề bài suy ra:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qamaaDa % aaleaacaWGUbaabaGaaGinaaaakiaaikdadaahaaWcbeqaaiaad6ga % cqGHsislcaaI0aaaaOGaeyypa0JaaGOnaiaaicdacqGHuhY2caWGdb % Waa0baaSqaaiaad6gaaeaacaaI0aaaaOGaaGOmamaaCaaaleqabaGa % amOBaaaakiabg2da9iaaiMdacaaI2aGaaGimaiaaykW7caaMc8+aae % WaaeaacaGGQaaacaGLOaGaayzkaaaaaa!4E36! C_n^4{2^{n - 4}} = 60 \Leftrightarrow C_n^4{2^n} = 960\,\,\left( * \right)\)

Tới đây ta dùng phương pháp thử trực tiếp đáp án và chỉ có n = 6 thỏa phương trình (*)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiqadg % eagaqbaiaab+cacaqGVaWaaeWaaeaacaWGcbGaam4qaiqadoeagaqb % aiqadkeagaqbaaGaayjkaiaawMcaaaaa!3DAE! AA'{\rm{//}}\left( {BCC'B'} \right)\) nên khoảng cách từ AA' đến mặt phẳng (BCC'B') cũng chính là khoảng cách từ A đến mặt phẳng (BCC'B').

Hạ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadI % eacqGHLkIxcaWGcbGaam4qaiabgkDiElaadgeacaWGibGaeyyPI41a % aeWaaeaacaWGcbGaam4qaiqadoeagaqbaiqadkeagaqbaaGaayjkai % aawMcaaaaa!4526! AH \bot BC \Rightarrow AH \bot \left( {BCC'B'} \right)\)

Ta có : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaamyqaiaadIeadaahaaWcbeqaaiaaikdaaaaaaOGaeyyp % a0ZaaSaaaeaacaaIXaaabaGaamyqaiaadkeadaahaaWcbeqaaiaaik % daaaaaaOGaey4kaSYaaSaaaeaacaaIXaaabaGaamyqaiaadoeadaah % aaWcbeqaaiaaikdaaaaaaOGaeyypa0ZaaSaaaeaacaaIXaaabaGaaG % 4maiaadggadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSYaaSaaaeaa % caaIXaaabaGaamOqaiaadoeadaahaaWcbeqaaiaaikdaaaGccqGHsi % slcaWGbbGaamOqamaaCaaaleqabaGaaGOmaaaaaaGccqGH9aqpdaWc % aaqaaiaaigdaaeaacaaIZaGaamyyamaaCaaaleqabaGaaGOmaaaaaa % GccqGHRaWkdaWcaaqaaiaaigdaaeaacaWGHbWaaWbaaSqabeaacaaI % Yaaaaaaakiabg2da9maalaaabaGaaGinaaqaaiaaiodacaWGHbWaaW % baaSqabeaacaaIYaaaaaaaaaa!5A13! \frac{1}{{A{H^2}}} = \frac{1}{{A{B^2}}} + \frac{1}{{A{C^2}}} = \frac{1}{{3{a^2}}} + \frac{1}{{B{C^2} - A{B^2}}} = \frac{1}{{3{a^2}}} + \frac{1}{{{a^2}}} = \frac{4}{{3{a^2}}}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yqaiaadIeacqGH9aqpdaWcaaqaaiaadggadaGcaaqaaiaaiodaaSqa % baaakeaacaaIYaaaaaaa!3D7D! \Rightarrow AH = \frac{{a\sqrt 3 }}{2}\)

Vậy khoảng cách từ (AA')  đến mặt phẳng (BCC'B') bằng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGHbWaaOaaaeaacaaIZaaaleqaaaGcbaGaaGOmaaaaaaa!3887! \frac{{a\sqrt 3 }}{2}\)

Theo đề bài, ta có : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qamaaDa % aaleaacaWGUbaabaGaaG4maaaakiabg2da9iaaikdacaWGdbWaa0ba % aSqaaiaad6gaaeaacaaIYaaaaaaa!3D09! C_n^3 = 2C_n^2\)(1) với  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOBaiabgw % MiZkaaiodaaaa!396A! n \ge 3\);\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOBaiabgI % Giohbbfv3ySLgzGueE0jxyaGqbaiab-vriLcaa!3E67! n \in R\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aaS % aaaeaacaWGUbGaaiyiaaqaaiaaiodacaGGHaWaaeWaaeaacaWGUbGa % eyOeI0IaaG4maaGaayjkaiaawMcaaiaacgcaaaGaeyypa0JaaGOmam % aalaaabaGaamOBaiaacgcaaeaacaaIYaGaaiyiamaabmaabaGaamOB % aiabgkHiTiaaikdaaiaawIcacaGLPaaacaGGHaaaaiabgsDiBpaala % aabaGaaGymaaqaaiaaiAdaaaGaeyypa0ZaaSaaaeaacaaIXaaabaGa % amOBaiabgkHiTiaaikdaaaGaeyi1HSTaamOBaiabg2da9iaaiIdaaa % a!5724! \Leftrightarrow \frac{{n!}}{{3!\left( {n - 3} \right)!}} = 2\frac{{n!}}{{2!\left( {n - 2} \right)!}} \Leftrightarrow \frac{1}{6} = \frac{1}{{n - 2}} \Leftrightarrow n = 8\)

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: \( y'\left( 1 \right) = \frac{1}{2} \Leftrightarrow \frac{e}{{e + {m^2}}} = \frac{1}{2} \Leftrightarrow 2e = e + {m^2} \Leftrightarrow m = \pm \sqrt e \)

Tập xác định: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiraiabg2 % da9maajadabaGaeyOeI0IaeyOhIuQaai4oaiaaigdaaiaawIcacaGL % DbaacqGHQicYdaqcsaqaaiaaiwdacaGG7aGaey4kaSIaeyOhIukaca % GLBbGaayzkaaaaaa!44D1! D = \left( { - \infty ;1} \right] \cup \left[ {5; + \infty } \right)\)

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0ZaaSaaaeaacaWG4bGaeyOeI0IaaG4maaqaamaakaaabaGa % amiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiAdacaWG4bGaey % 4kaSIaaGynaaWcbeaaaaGccqGH+aGpcaaIWaaaaa!42DD! y' = \frac{{x - 3}}{{\sqrt {{x^2} - 6x + 5} }} > 0\) ; \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiaIiIaam % iEaiabgIGiopaabmaabaGaaGynaiaacUdacqGHRaWkcqGHEisPaiaa % wIcacaGLPaaaaaa!3E9F! \forall x \in \left( {5; + \infty } \right)\)

Vậy hàm số đồng biến trên khoảng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aI1aGaai4oaiabgUcaRiabg6HiLcGaayjkaiaawMcaaiaac6caaaa!3C00! \left( {5; + \infty } \right).\)

Số cách chọn 4 học sinh lên bảng: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOBamaabm % aabaGaeyyQdCfacaGLOaGaayzkaaGaeyypa0Jaam4qamaaDaaaleaa % caaIZaGaaGynaaqaaiaaisdaaaaaaa!3E34! n\left( \Omega \right) = C_{35}^4\)

Số cách chọn 4 học sinh chỉ có nam hoặc chỉ có nữ: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qamaaDa % aaleaacaaIYaGaaGimaaqaaiaaisdaaaGccqGHRaWkcaWGdbWaa0ba % aSqaaiaaigdacaaI1aaabaGaaGinaaaaaaa!3D36! C_{20}^4 + C_{15}^4\)

Xác suất để 4 học sinh được gọi có cả nam và nữ:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiabgk % HiTmaalaaabaGaam4qamaaDaaaleaacaaIYaGaaGimaaqaaiaaisda % aaGccqGHRaWkcaWGdbWaa0baaSqaaiaaigdacaaI1aaabaGaaGinaa % aaaOqaaiaadoeadaqhaaWcbaGaaG4maiaaiwdaaeaacaaI0aaaaaaa % kiabg2da9maalaaabaGaaGinaiaaiAdacaaIXaGaaGynaaqaaiaaiw % dacaaIYaGaaG4maiaaiAdaaaaaaa!4937! 1 - \frac{{C_{20}^4 + C_{15}^4}}{{C_{35}^4}} = \frac{{4615}}{{5236}}\)

Xác suất để chọn được câu trả lời đúng là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaaGinaaaaaaa!377D! \frac{1}{4}\) , xác suất để chọn được câu trả lời sai là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIZaaabaGaaGinaaaaaaa!377F! \frac{3}{4}\)

Để được 6 điểm thì thí sinh đó phải trả lời đúng 30 câu và trả lời sai 20 câu.

Xác suất để thí sinh đó được 6 điểm là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qamaaDa % aaleaacaaI1aGaaGimaaqaaiaaikdacaaIWaaaaOWaaeWaaeaadaWc % aaqaaiaaiodaaeaacaaI0aaaaaGaayjkaiaawMcaamaaCaaaleqaba % GaaGOmaiaaicdaaaGcdaqadaqaamaalaaabaGaaGymaaqaaiaaisda % aaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIZaGaaGimaaaakiabg2 % da9iaaicdacaGGSaGaaGOmaiaaiwdadaahaaWcbeqaaiaaiodacaaI % WaaaaOGaaiOlaiaaicdacaGGSaGaaG4naiaaiwdadaahaaWcbeqaai % aaikdacaaIWaaaaOGaaiOlaiaadoeadaqhaaWcbaGaaGynaiaaicda % aeaacaaIYaGaaGimaaaaaaa!52DB! C_{50}^{20}{\left( {\frac{3}{4}} \right)^{20}}{\left( {\frac{1}{4}} \right)^{30}} = 0,{25^{30}}.0,{75^{20}}.C_{50}^{20}\)

Đồ thị (H) có tiệm cận đứng là x = 2

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaCbeaeaaci % GGSbGaaiyAaiaac2gaaSqaaiaadIhacqGHsgIRcqGHXcqScqGHEisP % aeqaaOGaamyEaiabg2da9maaxababaGaciiBaiaacMgacaGGTbaale % aacaWG4bGaeyOKH4QaeyySaeRaeyOhIukabeaakmaalaaabaGaaGOm % aiaaicdacaaIXaGaaG4naaqaaiaadIhacqGHsislcaaIYaaaaiabg2 % da9iaaicdacqGHshI3daqadaqaaiaadIeaaiaawIcacaGLPaaaaaa!56CA! \mathop {\lim }\limits_{x \to \pm \infty } y = \mathop {\lim }\limits_{x \to \pm \infty } \frac{{2017}}{{x - 2}} = 0 \Rightarrow \left( H \right)\) có tiệm cận ngang là y = 0

Vậy số đường tiệm cận của (H) là 2

Kẻ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4qayaafa % GaamisaiabgwQiEnaabmaabaGaamyqaiaadkeacaWGdbaacaGLOaGa % ayzkaaaaaa!3D23! C'H \bot \left( {ABC} \right)\) tại \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamisaiabgk % DiEpaaHaaabaWaaeWaaeaacaWGdbGabm4qayaafaGaai4oamaabmaa % baGaamyqaiaadkeacaWGdbaacaGLOaGaayzkaaaacaGLOaGaayzkaa % aacaGLcmaacqGH9aqpdaqiaaqaaiqadoeagaqbaiaadoeacaWGibaa % caGLcmaacaGGUaaaaa!4684! H \Rightarrow \widehat {\left( {CC';\left( {ABC} \right)} \right)} = \widehat {C'CH}.\)

Bài ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaada % qadaqaaiaadoeaceWGdbGbauaacaGG7aWaaeWaaeaacaWGbbGaamOq % aiaadoeaaiaawIcacaGLPaaaaiaawIcacaGLPaaaaiaawkWaaiabg2 % da9iaaiodacaaIWaGaeyiSaaRaeyO0H49aaecaaeaaceWGdbGbauaa % caWGdbGaamisaaGaayPadaGaeyypa0JaaG4maiaaicdacqGHWcaSaa % a!4CD1! \widehat {\left( {CC';\left( {ABC} \right)} \right)} = 30^\circ \Rightarrow \widehat {C'CH} = 30^\circ \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taci % 4CaiaacMgacaGGUbGaaG4maiaaicdacqGHWcaScqGH9aqpdaWcaaqa % aiqadoeagaqbaiaadIeaaeaacaWGdbGabm4qayaafaaaaiabg2da9m % aalaaabaGaaGymaaqaaiaaikdaaaGaeyO0H4Tabm4qayaafaGaamis % aiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaGaam4qaiqadoeaga % qbaiabg2da9maalaaabaGaaGOmamaakaaabaGaaG4maaWcbeaaaOqa % aiaaikdaaaGaeyypa0ZaaOaaaeaacaaIZaaaleqaaOGaaiOlaaaa!539C! \Rightarrow \sin 30^\circ = \frac{{C'H}}{{CC'}} = \frac{1}{2} \Rightarrow C'H = \frac{1}{2}CC' = \frac{{2\sqrt 3 }}{2} = \sqrt 3 .\)

Do đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGbbGaamOqaiaadoeacaGGUaGabmyqayaafaGabmOqayaa % faGabm4qayaafaaabeaakiabg2da9iqadoeagaqbaiaadIeacaGGUa % Gaam4uamaaBaaaleaacaWGbbGaamOqaiaadoeaaeqaaaaa!4336! {V_{ABC.A'B'C'}} = C'H.{S_{ABC}}\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0Jabm % 4qayaafaGaamisaiaac6cadaWcaaqaaiaaigdaaeaacaaIYaaaaiaa % dgeacaWGcbGaaiOlaiaadgeacaWGdbGaaiOlaiGacohacaGGPbGaai % OBaiaaiAdacaaIWaGaeyiSaaRaeyypa0ZaaOaaaeaacaaIZaaaleqa % aOGaaiOlamaalaaabaGaaGymaaqaaiaaikdaaaGaaiOlaiaaiodaca % GGUaGaaG4maiaac6cadaWcaaqaamaakaaabaGaaG4maaWcbeaaaOqa % aiaaikdaaaGaeyypa0ZaaSaaaeaacaaIYaGaaG4naaqaaiaaisdaaa % aaaa!5240! = C'H.\frac{1}{2}AB.AC.\sin 60^\circ = \sqrt 3 .\frac{1}{2}.3.3.\frac{{\sqrt 3 }}{2} = \frac{{27}}{4}\)

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGtbGaaiOlaiaadkeacaWGdbGaamiraaqabaGccqGH9aqp % daWcaaqaaiaaigdaaeaacaaIZaaaaiaadofacaWGbbGaaiOlaiaado % fadaWgaaWcbaGaamOqaiaadoeacaWGebaabeaakiaac6caaaa!43DC! {V_{S.BCD}} = \frac{1}{3}SA.{S_{BCD}}.\)

Lại có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaWGcbGaam4qaiaadseaaeqaaOGaeyypa0Jaam4uamaaBaaa % leaacaWGbbGaamOqaiaadoeacaWGebaabeaakiabgkHiTiaadofada % WgaaWcbaGaamyqaiaadkeacaWGebaabeaaaaa!42D2! {S_{BCD}} = {S_{ABCD}} - {S_{ABD}}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaaIXaaabaGaaGOmaaaacaWGbbGaamOqaiaac6cadaqadaqa % aiaadgeacaWGebGaey4kaSIaamOqaiaadoeaaiaawIcacaGLPaaacq % GHsisldaWcaaqaaiaaigdaaeaacaaIYaaaaiaadgeacaWGcbGaaiOl % aiaadgeacaWGebaaaa!468A! = \frac{1}{2}AB.\left( {AD + BC} \right) - \frac{1}{2}AB.AD\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaaIXaaabaGaaGOmaaaacaWGbbGaamOqaiaac6cacaWGcbGa % am4qaiabg2da9maalaaabaGaaGymaaqaaiaaikdaaaGaamyyamaaCa % aaleqabaGaaGOmaaaakiaac6caaaa!4166! = \frac{1}{2}AB.BC = \frac{1}{2}{a^2}.\)

Mà \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaadg % eacqGH9aqpcaWGHbWaaOaaaeaacaaIZaaaleqaaOGaeyO0H4TaamOv % amaaBaaaleaacaWGtbGaaiOlaiaadkeacaWGdbGaamiraaqabaGccq % GH9aqpdaWcaaqaaiaaigdaaeaacaaIZaaaaiaadggadaGcaaqaaiaa % iodaaSqabaGccaGGUaWaaSaaaeaacaWGHbWaaWbaaSqabeaacaaIYa % aaaaGcbaGaaGOmaaaacqGH9aqpdaWcaaqaaiaadggadaahaaWcbeqa % aiaaiodaaaGcdaGcaaqaaiaaiodaaSqabaaakeaacaaI2aaaaiaac6 % caaaa!4EA0! SA = a\sqrt 3 \Rightarrow {V_{S.BCD}} = \frac{1}{3}a\sqrt 3 .\frac{{{a^2}}}{2} = \frac{{{a^3}\sqrt 3 }}{6}.\)

Nhận xét: Nếu đề bài bỏ giả thiết AD = 3a thì sẽ giải như sau:

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGtbGaaiOlaiaadkeacaWGdbGaamiraaqabaGccqGH9aqp % daWcaaqaaiaaigdaaeaacaaIZaaaaiaadofacaWGbbGaaiOlaiaado % fadaWgaaWcbaGaamOqaiaadoeacaWGebaabeaakiabg2da9maalaaa % baGaaGymaaqaaiaaiodaaaGaam4uaiaadgeacaGGUaWaaSaaaeaaca % aIXaaabaGaaGOmaaaacaWGKbWaaeWaaeaacaWGebGaaiilaiaadkea % caWGdbaacaGLOaGaayzkaaGaaiOlaiaadkeacaWGdbaaaa!514A! {V_{S.BCD}} = \frac{1}{3}SA.{S_{BCD}} = \frac{1}{3}SA.\frac{1}{2}d\left( {D,BC} \right).BC\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaS % aaaeaacaaIXaaabaGaaGOnaaaacaWGtbGaamyqaiaac6cacaWGbbGa % amOqaiaac6cacaWGcbGaam4qaiabg2da9maalaaabaGaamyyamaaCa % aaleqabaGaaG4maaaakmaakaaabaGaaG4maaWcbeaaaOqaaiaaiAda % aaaaaa!4334! = \frac{1}{6}SA.AB.BC = \frac{{{a^3}\sqrt 3 }}{6}\)

Thể tích: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9maalaaabaGaaGymaaqaaiaaiodaaaGaeqiWdaNaamOuamaaCaaa % leqabaGaaGOmaaaakiaadIgacqGH9aqpdaWcaaqaaiaaigdaaeaaca % aIZaaaaiabec8aWjaac6cacaWGpbGaamyqamaaCaaaleqabaGaaGOm % aaaakiaac6cacaWGtbGaam4taiaac6caaaa!486A! V = \frac{1}{3}\pi {R^2}h = \frac{1}{3}\pi .O{A^2}.SO.\)

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGbbGaam4uaiaadkeaaiaawkWaaiabg2da9iaaiAdacaaIWaGaeyiS % aaRaeyO0H49aaecaaeaacaWGbbGaam4uaiaad+eaaiaawkWaaiabg2 % da9iaaiodacaaIWaGaeyiSaalaaa!4780! \widehat {ASB} = 60^\circ \Rightarrow \widehat {ASO} = 30^\circ \)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taci % iDaiaacggacaGGUbGaaG4maiaaicdacqGHWcaScqGH9aqpdaWcaaqa % aiaad+eacaWGbbaabaGaam4uaiaad+eaaaGaeyypa0ZaaSaaaeaaca % aIXaaabaWaaOaaaeaacaaIZaaaleqaaaaakiabgkDiElaadofacaWG % pbGaeyypa0Jaam4taiaadgeadaGcaaqaaiaaiodaaSqabaGccaGGUa % aaaa!4DD0! \Rightarrow \tan 30^\circ = \frac{{OA}}{{SO}} = \frac{1}{{\sqrt 3 }} \Rightarrow SO = OA\sqrt 3 .\)

Lại có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaWG4bGaamyCaaqabaGccqGH9aqpcqaHapaCcaWGsbGaamiB % aiabg2da9iabec8aWjaac6cacaWGpbGaamyqaiaac6cacaWGtbGaam % yqaiabg2da9iabec8aWjaac6cacaWGpbGaamyqamaakaaabaGaam4t % aiaadgeadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWGtbGaam4tam % aaCaaaleqabaGaaGOmaaaaaeqaaOGaeyypa0JaaGOnaiabec8aWjaa % dggadaahaaWcbeqaaiaaikdaaaaaaa!555D! {S_{xq}} = \pi Rl = \pi .OA.SA = \pi .OA\sqrt {O{A^2} + S{O^2}} = 6\pi {a^2}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % 4taiaadgeadaGcaaqaaiaad+eacaWGbbWaaWbaaSqabeaacaaIYaaa % aOGaey4kaSIaaG4maiaad+eacaWGbbWaaWbaaSqabeaacaaIYaaaaa % qabaGccqGH9aqpcaaI2aGaamyyamaaCaaaleqabaGaaGOmaaaakiab % gkDiElaaikdacaWGpbGaamyqamaaCaaaleqabaGaaGOmaaaakiabg2 % da9iaaiAdacaWGHbWaaWbaaSqabeaacaaIYaaaaaaa!4D8D! \Rightarrow OA\sqrt {O{A^2} + 3O{A^2}} = 6{a^2} \Rightarrow 2O{A^2} = 6{a^2}\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % 4taiaadgeacqGH9aqpcaWGHbWaaOaaaeaacaaIZaaaleqaaOGaeyO0 % H4Taam4uaiaad+eacqGH9aqpcaaIZaGaamyyaiabgkDiElaadAfacq % GH9aqpdaWcaaqaaiaaigdaaeaacaaIZaaaaiabec8aWjaac6cacaaI % ZaGaamyyamaaCaaaleqabaGaaGOmaaaakiaac6cacaaIZaGaamyyai % abg2da9iaaiodacqaHapaCcaWGHbWaaWbaaSqabeaacaaIZaaaaOGa % aiOlaaaa!5696! \Rightarrow OA = a\sqrt 3 \Rightarrow SO = 3a \Rightarrow V = \frac{1}{3}\pi .3{a^2}.3a = 3\pi {a^3}.\)

Ta có ngay kết quả sau:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGbbGaam4qaiaadkeacaGGNaGaamiraiaacEcaaeqaaOGa % eyypa0JaamOvaiabgkHiTmaabmaabaGaamOvamaaBaaaleaacaWGcb % Gaai4jaiaac6cacaWGbbGaamOqaiaadoeaaeqaaOGaey4kaSIaamOv % amaaBaaaleaacaWGdbGaaiOlaiaadkeacaGGNaGaam4qaiaacEcaca % WGebGaai4jaaqabaGccqGHRaWkcaWGwbWaaSbaaSqaaiaadseacaGG % NaGaaiOlaiaadgeacaWGdbGaamiraaqabaGccqGHRaWkcaWGwbWaaS % baaSqaaiaadgeacaGGUaGaamyqaiaacEcacaWGcbGaai4jaiaadsea % caGGNaaabeaaaOGaayjkaiaawMcaaiaac6caaaa!5C03! {V_{ACB'D'}} = V - \left( {{V_{B'.ABC}} + {V_{C.B'C'D'}} + {V_{D'.ACD}} + {V_{A.A'B'D'}}} \right).\)

Lưu ý: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGcbGaai4jaiaac6cacaWGbbGaamOqaiaadoeaaeqaaOGa % eyypa0JaamOvamaaBaaaleaacaWGdbGaaiOlaiaadkeacaGGNaGaam % 4qaiaacEcacaWGebGaai4jaaqabaGccqGH9aqpcaWGwbWaaSbaaSqa % aiaadseacaGGNaGaaiOlaiaadgeacaWGdbGaamiraaqabaGccqGH9a % qpcaWGwbWaaSbaaSqaaiaadgeacaGGUaGaamyqaiaacEcacaWGcbGa % ai4jaiaadseacaGGNaaabeaakiabg2da9maalaaabaGaaGymaaqaai % aaiodaaaGaamOvamaaBaaaleaacaWGbbGaamOqaiaadoeacaGGUaGa % amyqaiaacEcacaWGcbGaai4jaiaadoeacaGGNaaabeaakiabg2da9m % aalaaabaGaaGymaaqaaiaaiodaaaGaaiOlamaalaaabaGaamOvaaqa % aiaaikdaaaGaeyO0H4TaamOvamaaBaaaleaacaWGbbGaam4qaiaadk % eacaGGNaGaamiraiaacEcaaeqaaOGaeyypa0JaamOvaiabgkHiTiaa % isdacaGGUaWaaSaaaeaacaWGwbaabaGaaGOnaaaacqGH9aqpdaWcaa % qaaiaadAfaaeaacaaIZaaaaiaac6caaaa!72EF! {V_{B'.ABC}} = {V_{C.B'C'D'}} = {V_{D'.ACD}} = {V_{A.A'B'D'}} = \frac{1}{3}{V_{ABC.A'B'C'}} = \frac{1}{3}.\frac{V}{2} \Rightarrow {V_{ACB'D'}} = V - 4.\frac{V}{6} = \frac{V}{3}.\)

Gọi I;I' lần lượt là tâm hai đáy,  O là trung điểm của II'. Khi đó ta có O là tâm mặt cầu ngoại tiếp lăng trụ.

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadM % eacqGH9aqpdaWcaaqaaiaadggadaGcaaqaaiaaiodaaSqabaaakeaa % caaIZaaaaiaacYcacaaMc8Uaamysaiaad+eacqGH9aqpdaWcaaqaai % aadkgaaeaacaaIYaaaaaaa!41B9! AI = \frac{{a\sqrt 3 }}{3},\,IO = \frac{b}{2}\) suy ra bán kính mặt cầu ngoại tiếp lăng trụ là  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabg2 % da9maakaaabaWaaSaaaeaacaWGHbWaaWbaaSqabeaacaaIYaaaaaGc % baGaaG4maaaacqGHRaWkdaWcaaqaaiaadkgadaahaaWcbeqaaiaaik % daaaaakeaacaaI0aaaaaWcbeaakiabg2da9maalaaabaGaaGymaaqa % aiaaikdadaGcaaqaaiaaiodaaSqabaaaaOWaaOaaaeaacaaI0aGaam % yyamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiodacaWGIbWaaWba % aSqabeaacaaIYaaaaaqabaaaaa!47AA! R = \sqrt {\frac{{{a^2}}}{3} + \frac{{{b^2}}}{4}} = \frac{1}{{2\sqrt 3 }}\sqrt {4{a^2} + 3{b^2}} \)

Vậy: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaadaqadaqaaiaad+eacaGG7aGaaGPaVlaadkfaaiaawIcacaGL % PaaaaeqaaOGaeyypa0ZaaSaaaeaacaaI0aaabaGaaG4maaaacqaHap % aCcaWGsbWaaWbaaSqabeaacaaIZaaaaOGaeyypa0ZaaSaaaeaacqaH % apaCaeaacaaIXaGaaGioamaakaaabaGaaG4maaWcbeaaaaGcdaGcaa % qaamaabmaabaGaaGinaiaadggadaahaaWcbeqaaiaaikdaaaGccqGH % RaWkcaaIZaGaamOyamaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawM % caamaaCaaaleqabaGaaG4maaaaaeqaaOGaaiOlaaaa!511D! {V_{\left( {O;\,R} \right)}} = \frac{4}{3}\pi {R^3} = \frac{\pi }{{18\sqrt 3 }}\sqrt {{{\left( {4{a^2} + 3{b^2}} \right)}^3}} .\)

Ta có:\(\Delta MAB\)  vuông tại M có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGcbaacaGLcmaacqGH9aqpcaaI2aGaaGimaiabgclaWcaa!3BE9! \widehat B = 60^\circ \) nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiaadk % eacqGH9aqpdaGcaaqaaiaaiodaaSqabaGccaGG7aaaaa!3A34! MB = \sqrt 3 ;\)MA = 3

Gọi H là hình chiếu của M lên AB , suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiaadI % eacqGHLkIxdaqadaqaaiaadgeacaWGdbGaamiraaGaayjkaiaawMca % aaaa!3D24! MH \bot \left( {ACD} \right)\) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiaadI % eacqGH9aqpdaWcaaqaaiaad2eacaWGcbGaaiOlaiaad2eacaWGbbaa % baGaamyqaiaadkeaaaGaeyypa0ZaaSaaaeaacaaIZaaabaGaaGOmaa % aacaGGUaaaaa!415A! MH = \frac{{MB.MA}}{{AB}} = \frac{3}{2}.\)

Vậy \( {V_{M.ACD}} = \frac{1}{3}MH.S{ _{ACD}} = \frac{1}{3}.\frac{3}{2}.6 = 3\,\left( {{\rm{c}}{{\rm{m}}^{\rm{3}}}} \right)\)

Điều kiện:  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaCa % aaleqabaGaaGOmaaaakiabgkHiTiaaikdacaWGTbGaamiEaiabgUca % RiaaisdacqGH+aGpcaaIWaGaaGPaVlaaykW7caaMc8UaaGPaVpaabm % aabaGaaiOkaaGaayjkaiaawMcaaaaa!4741! {x^2} - 2mx + 4 > 0\,\,\,\,\left( * \right)\)

Để (*) đúng với mọi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % Giolabl2riHcaa!39E5! x \in R\) thì  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGafuiLdqKbau % aacqGH9aqpcaWGTbWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGin % aiabgYda8iaaicdacqGHuhY2cqGHsislcaaIYaGaeyipaWJaamyBai % abgYda8iaaikdaaaa!4575! \Delta ' = {m^2} - 4 < 0 \Leftrightarrow - 2 < m < 2\)

Theo bài ra ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaad+ % eacqGH9aqpcaWGYbGaeyypa0JaaGOmaiaaiwdacaGG7aaaaa!3CCB! AO = r = 25;OK=12\);  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaad+ % eacqGH9aqpcaWGObGaeyypa0JaaGOmaiaaicdacaGG7aGaaGPaVdaa % !3E59! SO = h = 20;\)(Hình vẽ).

Lại có  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaam4taiaadUeadaahaaWcbeqaaiaaikdaaaaaaOGaeyyp % a0ZaaSaaaeaacaaIXaaabaGaam4taiaadMeadaahaaWcbeqaaiaaik % daaaaaaOGaey4kaSYaaSaaaeaacaaIXaaabaGaam4taiaadofadaah % aaWcbeqaaiaaikdaaaaaaOGaeyO0H4Taam4taiaadMeacqGH9aqpca % aIXaGaaGynaiaaykW7daqadaqaaiaabogacaqGTbaacaGLOaGaayzk % aaaaaa!4D71! \frac{1}{{O{K^2}}} = \frac{1}{{O{I^2}}} + \frac{1}{{O{S^2}}} \Rightarrow OI = 15\,\left( {{\rm{cm}}} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadk % eacqGH9aqpcaaIYaGaamyqaiaadMeacqGH9aqpdaGcaaqaaiaaikda % caaI1aWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGymaiaaiwdada % ahaaWcbeqaaiaaikdaaaaabeaakiabg2da9iaaisdacaaIWaGaaGPa % VpaabmaabaGaae4yaiaab2gaaiaawIcacaGLPaaacaGG7aGaaGjbVl % aadofacaWGjbGaeyypa0ZaaOaaaeaacaWGtbGaam4tamaaCaaaleqa % baGaaGOmaaaakiabgUcaRiaad+eacaWGjbWaaWbaaSqabeaacaaIYa % aaaaqabaGccqGH9aqpcaaIYaGaaGynaiaaykW7daqadaqaaiaaboga % caqGTbaacaGLOaGaayzkaaGaeyO0H4Taam4uamaaBaaaleaacqqHuo % arcaWGtbGaamyqaiaadkeaaeqaaOGaeyypa0ZaaSaaaeaacaaIXaaa % baGaaGOmaaaacaGGUaGaaGOmaiaaiwdacaGGUaGaaGinaiaaicdacq % GH9aqpcaaI1aGaaGimaiaaicdacaaMc8+aaeWaaeaacaqGJbGaaeyB % amaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaaiaac6caaaa!7387! AB = 2AI = \sqrt {{{25}^2} - {{15}^2}} = 40\,\left( {{\rm{cm}}} \right);\;SI = \sqrt {S{O^2} + O{I^2}} = 25\,\left( {{\rm{cm}}} \right) \Rightarrow {S_{\Delta SAB}} = \frac{1}{2}.25.40 = 500\,\left( {{\rm{c}}{{\rm{m}}^2}} \right).\)

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 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadggadaahaaWcbeqaaiaadIhaaaGccaGGSaGaaGPaVlaadMha % cqGH9aqpcaWGIbWaaWbaaSqabeaacaWG4baaaaaa!4062! y = {a^x},\,y = {b^x}\) đồng biến nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabg6 % da+iaaigdacaGG7aGaaGPaVlaadkgacqGH+aGpcaaIXaaaaa!3D91! a > 1;\,b > 1\) nên c là số nhỏ nhất trong ba số.

Đường thẳng x = 1 cắt hai hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadggadaahaaWcbeqaaiaadIhaaaGccaGGSaaaaa!3AC2! y = {a^x},\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadkgadaahaaWcbeqaaiaadIhaaaaaaa!3A09! y = {b^x}\) tại các điểm có tung độ lần lượt là a và b, dễ thấy a > b (hình vẽ). Vậy c < b < a

Gọi D là hình chiếu của S lên mặt phẳng (ABC) , suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaads % eacqGHLkIxdaqadaqaaiaadgeacaWGcbGaam4qaaGaayjkaiaawMca % aaaa!3D24! SD \bot \left( {ABC} \right)\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaads % eacqGHLkIxcaWGbbGaamOqaaaa!3AD3! SD \bot AB\) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaadk % eacqGHLkIxcaWGbbGaamOqaiaaykW7caGGOaGaam4zaiaadshacaGG % Paaaaa!3F9A! SB \bot AB\,(gt)\), suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadk % eacqGHLkIxdaqadaqaaiaadofacaWGcbGaamiraaGaayjkaiaawMca % aiabgkDiElaadkeacaWGbbGaeyyPI4LaamOqaiaadseaaaa!444E! AB \bot \left( {SBD} \right) \Rightarrow BA \bot BD\)

Tương tự \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaado % eacqGHLkIxcaWGebGaam4qaaaa!3AC4! AC \bot DC\) có  hay tam giác ACD vuông ở C.

Dễ thấy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam % 4uaiaadkeacaWGbbGaeyypa0JaeuiLdqKaam4uaiaadoeacaWGbbaa % aa!3E91! \Delta SBA = \Delta SCA\) (cạnh huyền và cạnh góc vuông), suy ra SB=SC . Từ đó ta chứng minh được \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeuiLdqKaam % 4uaiaadkeacaWGebGaeyypa0JaeuiLdqKaam4uaiaadoeacaWGebaa % aa!3E97! \Delta SBD = \Delta SCD\) nên cũng có DB=DC.

Vậy DA là đường trung trực của BC , nên cũng là đường phân giác của góc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGcbGaamyqaiaadoeaaiaawkWaaaaa!390B! \widehat {BAC}\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGebGaamyqaiaadoeaaiaawkWaaiabg2da9iaaiodacaaIWaGaeyiS % aalaaa!3D76! \widehat {DAC} = 30^\circ \), suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiraiaado % eacqGH9aqpdaWcaaqaaiaadggaaeaadaGcaaqaaiaaiodaaSqabaaa % aaaa!3A59! DC = \frac{a}{{\sqrt 3 }}\) . Ngoài ra góc giữa hai mặt phẳng (SAB) và (ABC) là  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGtbGaamOqaiaadseaaiaawkWaaiabg2da9iaaiAdacaaIWaGaeyiS % aalaaa!3D8A! \widehat {SBD} = 60^\circ\), suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiDaiaacg % gacaGGUbWaaecaaeaacaWGtbGaamOqaiaadseaaiaawkWaaiabg2da % 9maalaaabaGaam4uaiaadseaaeaacaWGcbGaamiraaaacqGHshI3ca % WGtbGaamiraiabg2da9iaadkeacaWGebGaciiDaiaacggacaGGUbWa % aecaaeaacaWGtbGaamOqaiaadseaaiaawkWaaiabg2da9maalaaaba % GaamyyaaqaamaakaaabaGaaG4maaWcbeaaaaGccaGGUaWaaOaaaeaa % caaIZaaaleqaaOGaeyypa0Jaamyyaaaa!5323! \tan \widehat {SBD} = \frac{{SD}}{{BD}} \Rightarrow SD = BD\tan \widehat {SBD} = \frac{a}{{\sqrt 3 }}.\sqrt 3 = a\)

Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGtbGaaiOlaiaadgeacaWGcbGaam4qaaqabaGccqGH9aqp % daWcaaqaaiaaigdaaeaacaaIZaaaaiaac6cacaWGtbWaaSbaaSqaai % abfs5aejaadgeacaWGcbGaam4qaaqabaGccaGGUaGaam4uaiaadsea % cqGH9aqpdaWcaaqaaiaaigdaaeaacaaIZaaaaiaac6cadaWcaaqaai % aadggadaahaaWcbeqaaiaaikdaaaGcdaGcaaqaaiaaiodaaSqabaaa % keaacaaI0aaaaiaac6cacaWGHbGaeyypa0ZaaSaaaeaacaWGHbWaaW % baaSqabeaacaaIZaaaaOWaaOaaaeaacaaIZaaaleqaaaGcbaGaaGym % aiaaikdaaaaaaa!52EA! {V_{S.ABC}} = \frac{1}{3}.{S_{\Delta ABC}}.SD = \frac{1}{3}.\frac{{{a^2}\sqrt 3 }}{4}.a = \frac{{{a^3}\sqrt 3 }}{{12}}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaamaakaaabaGaaGOmaaadbeaaaSqabaGcdaqa % daqaaiaadIhacqGHsislcaaIXaaacaGLOaGaayzkaaGaeyypa0Jaci % iBaiaac+gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWG % TbGaamiEaiabgkHiTiaaiIdaaiaawIcacaGLPaaacqGHuhY2caaMc8 % UaaGPaVpaaceaaeaqabeaacaWG4bGaeyOpa4JaaGymaaqaamaabmaa % baGaamiEaiabgkHiTiaaigdaaiaawIcacaGLPaaadaahaaWcbeqaai % aaikdaaaGccqGH9aqpcaWGTbGaamiEaiabgkHiTiaaiIdaaaGaay5E % aaGaaGPaVlaaykW7cqGHuhY2caaMc8UaaGPaVpaaceaaeaqabeaaca % WG4bGaeyOpa4JaaGymaaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGc % cqGHsisldaqadaqaaiaad2gacqGHRaWkcaaIYaaacaGLOaGaayzkaa % GaamiEaiabgUcaRiaaiMdacqGH9aqpcaaIWaaaaiaawUhaaiaaykW7 % caaMc8oaaa!75E8! {\log _{\sqrt 2 }}\left( {x - 1} \right) = {\log _2}\left( {mx - 8} \right) \Leftrightarrow \,\,\left\{ \begin{array}{l} x > 1\\ {\left( {x - 1} \right)^2} = mx - 8 \end{array} \right.\,\, \Leftrightarrow \,\,\left\{ \begin{array}{l} x > 1\\ {x^2} - \left( {m + 2} \right)x + 9 = 0 \end{array} \right.\,\,\)

Để phương trình đã cho có hai nghiệm thực lớn hơn 1 thì điều kiện sau thỏa mãn.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGPaVlaayk % W7daGabaabaeqabaGaeuiLdqKaeyOpa4JaaGimaaqaaiaaigdacqGH % 8aapcaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaeyipaWJaamiEamaaBa % aaleaacaaIYaaabeaaaaGccaGL7baacaaMc8UaaGPaVlabgsDiBlaa % ykW7daGabaabaeqabaGaamyBamaaCaaaleqabaGaaGOmaaaakiabgU % caRiaaisdacaWGTbGaeyOeI0IaaG4maiaaikdacqGH+aGpcaaIWaaa % baWaaeWaaeaacaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaeyOeI0IaaG % ymaaGaayjkaiaawMcaaiabgUcaRmaabmaabaGaamiEamaaBaaaleaa % caaIYaaabeaakiabgkHiTiaaigdaaiaawIcacaGLPaaacqGH+aGpca % aIWaaabaWaaeWaaeaacaWG4bWaaSbaaSqaaiaaigdaaeqaaOGaeyOe % I0IaaGymaaGaayjkaiaawMcaamaabmaabaGaamiEamaaBaaaleaaca % aIYaaabeaakiabgkHiTiaaigdaaiaawIcacaGLPaaacqGH+aGpcaaI % WaaaaiaawUhaaiaaykW7caaMc8Uaeyi1HS9aaiqaaqaabeqaamaade % aaeaqabeaacaWGTbGaeyipaWJaeyOeI0IaaGioaaqaaiaad2gacqGH % +aGpcaaI0aaaaiaawUfaaaqaaiaad2gacqGH+aGpcaaIWaaabaGaaG % ioaiabgkHiTiaad2gacqGH+aGpcaaIWaaaaiaawUhaaiaaykW7caaM % c8Uaeyi1HSTaaGinaiabgYda8iaad2gacqGH8aapcaaI4aaaaa!8C8A! \,\,\left\{ \begin{array}{l} \Delta > 0\\ 1 < {x_1} < {x_2} \end{array} \right.\,\, \Leftrightarrow \,\left\{ \begin{array}{l} {m^2} + 4m - 32 > 0\\ \left( {{x_1} - 1} \right) + \left( {{x_2} - 1} \right) > 0\\ \left( {{x_1} - 1} \right)\left( {{x_2} - 1} \right) > 0 \end{array} \right.\,\, \Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} m < - 8\\ m > 4 \end{array} \right.\\ m > 0\\ 8 - m > 0 \end{array} \right.\,\, \Leftrightarrow 4 < m < 8\)

Vì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabgI % GiolablssiIkabgkDiElaad2gacqGHiiIZdaGadaqaaiaaiwdacaGG % SaGaaGOnaiaacYcacaaI3aaacaGL7bGaayzFaaaaaa!4486! m \in Z\Rightarrow m \in \left\{ {5,6,7} \right\}\)

Ta có tam giác ABC vuông tại A góc \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGbbGaamOqaiaadoeaaiaawkWaaiabg2da9iaaiodacaaIWaGaeyiS % aalaaa!3D74! \widehat {ABC} = 30^\circ\) và BC = a, suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaado % eacqGH9aqpdaWcaaqaaiaadggaaeaacaaIYaaaaiaacYcacaaMc8Ua % amyqaiaadkeacqGH9aqpdaWcaaqaaiaadggadaGcaaqaaiaaiodaaS % qabaaakeaacaaIYaaaaaaa!419C! AC = \frac{a}{2},\,AB = \frac{{a\sqrt 3 }}{2}\) .

Lại có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGPaVlaayk % W7daGabaabaeqabaWaaeWaaeaacaWGtbGaamyqaiaadkeaaiaawIca % caGLPaaacqGHLkIxdaqadaqaaiaadgeacaWGcbGaam4qaaGaayjkai % aawMcaaaqaaiaadoeacaWGbbGaeyyPI4LaamyqaiaadkeaaaGaay5E % aaGaaGPaVlaaykW7cqGHshI3caWGbbGaam4qaiabgwQiEnaabmaaba % Gaam4uaiaadgeacaWGcbaacaGLOaGaayzkaaaaaa!5513! \,\,\left\{ \begin{array}{l} \left( {SAB} \right) \bot \left( {ABC} \right)\\ CA \bot AB \end{array} \right.\,\, \Rightarrow AC \bot \left( {SAB} \right)\), suy ra tam giác SAC vuông tại A .

Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaadg % eacqGH9aqpdaGcaaqaaiaadofacaWGdbWaaWbaaSqabeaacaaIYaaa % aOGaeyOeI0IaamyqaiaadoeadaahaaWcbeqaaiaaikdaaaaabeaaki % abg2da9maakaaabaGaamyyamaaCaaaleqabaGaaGOmaaaakiabgkHi % TmaabmaabaWaaSaaaeaacaWGHbaabaGaaGOmaaaaaiaawIcacaGLPa % aadaahaaWcbeqaaiaaikdaaaaabeaakiabg2da9maalaaabaGaamyy % amaakaaabaGaaG4maaWcbeaaaOqaaiaaikdaaaaaaa!4A4D! SA = \sqrt {S{C^2} - A{C^2}} = \sqrt {{a^2} - {{\left( {\frac{a}{2}} \right)}^2}} = \frac{{a\sqrt 3 }}{2}\)

Tam giác SAB có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaadg % eacqGH9aqpdaWcaaqaaiaadggadaGcaaqaaiaaiodaaSqabaaakeaa % caaIYaaaaiaacYcacaaMc8UaamyqaiaadkeacqGH9aqpdaWcaaqaai % aadggadaGcaaqaaiaaiodaaSqabaaakeaacaaIYaaaaiaacYcacaaM % b8UaaGPaVlaadofacaWGcbGaeyypa0Jaamyyaaaa!49DE! SA = \frac{{a\sqrt 3 }}{2},\,AB = \frac{{a\sqrt 3 }}{2},\,SB = a\) . Từ đó sử dụng công thức Hê-rông ta tính được .\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaWGtbGaamyqaiaadkeaaeqaaOGaeyypa0ZaaSaaaeaacaWG % HbWaaWbaaSqabeaacaaIYaaaaOWaaOaaaeaacaaIYaaaleqaaaGcba % GaaGinaaaacqGHshI3caWGtbGaamisaiabg2da9maalaaabaGaaGOm % aiaadofadaWgaaWcbaGaam4uaiaadgeacaWGcbaabeaaaOqaaiaadg % eacaWGcbaaaiabg2da9maalaaabaGaamyyamaakaaabaGaaGOnaaWc % beaaaOqaaiaaiodaaaGaeyO0H4TaamOqaiaadIeacqGH9aqpdaWcaa % qaaiaadggadaGcaaqaaiaaiodaaSqabaaakeaacaaIZaaaaiabg2da % 9maalaaabaGaaGOmaiaadgeacaWGcbaabaGaaG4maaaaaaa!580F! {S_{SAB}} = \frac{{{a^2}\sqrt 2 }}{4} \Rightarrow SH = \frac{{2{S_{SAB}}}}{{AB}} = \frac{{a\sqrt 6 }}{3} \Rightarrow BH = \frac{{a\sqrt 3 }}{3} = \frac{{2AB}}{3}\)

Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamisaiaacYcadaqadaqaaiaadofacaWGcbGaam4qaaGaayjk % aiaawMcaaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaGOmaaqaai % aaiodaaaGaamizamaabmaabaGaamyqaiaacYcadaqadaqaaiaadofa % caWGcbGaam4qaaGaayjkaiaawMcaaaGaayjkaiaawMcaaiaac6caaa % a!48EC! d\left( {H,\left( {SBC} \right)} \right) = \frac{2}{3}d\left( {A,\left( {SBC} \right)} \right).\)Từ H kẻ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamisaiaadU % eacqGHLkIxcaWGcbGaam4qaaaa!3AD1! HK \bot BC\)

Kẻ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamisaiaadw % eacqGHLkIxcaWGtbGaam4saiabgkDiElaadIeacaWGfbGaeyyPI41a % aeWaaeaacaWGtbGaamOqaiaadoeaaiaawIcacaGLPaaaaaa!4479! HE \bot SK \Rightarrow HE \bot \left( {SBC} \right)\). Ta dễ tính được \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamisaiaadU % eacqGH9aqpdaWcaaqaaiaadggadaGcaaqaaiaaiodaaSqabaaakeaa % caaI2aaaaiabgkDiElaadsgadaqadaqaaiaadIeacaGGSaWaaeWaae % aacaWGtbGaamOqaiaadoeaaiaawIcacaGLPaaaaiaawIcacaGLPaaa % cqGH9aqpdaWcaaqaaiaadggadaGcaaqaaiaaiAdaaSqabaaakeaaca % aI5aaaaiaac6caaaa!49C1! HK = \frac{{a\sqrt 3 }}{6} \Rightarrow d\left( {H,\left( {SBC} \right)} \right) = \frac{{a\sqrt 6 }}{9}.\)

Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamyqaiaacYcadaqadaqaaiaadofacaWGcbGaam4qaaGaayjk % aiaawMcaaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaG4maaqaai % aaikdaaaGaamizamaabmaabaGaamisaiaacYcadaqadaqaaiaadofa % caWGcbGaam4qaaGaayjkaiaawMcaaaGaayjkaiaawMcaaiabg2da9m % aalaaabaGaaG4maaqaaiaaikdaaaGaeyyXIC9aaSaaaeaacaWGHbWa % aOaaaeaacaaI2aaaleqaaaGcbaGaaGyoaaaacqGH9aqpdaWcaaqaai % aadggadaGcaaqaaiaaiAdaaSqabaaakeaacaaI2aaaaaaa!5352! d\left( {A,\left( {SBC} \right)} \right) = \frac{3}{2}d\left( {H,\left( {SBC} \right)} \right) = \frac{3}{2} \cdot \frac{{a\sqrt 6 }}{9} = \frac{{a\sqrt 6 }}{6}\)

Gọi  là trung điểm OA . Vì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiaad2 % eacaqGVaGaae4laiaadofacaWGpbGaeyO0H4Taamysaiaad2eacqGH % LkIxdaqadaqaaiaadgeacaWGcbGaam4qaiaadseaaiaawIcacaGLPa % aaaaa!44F9! IM{\rm{//}}SO \Rightarrow IM \bot \left( {ABCD} \right)\) nên hình chiếu của MN lên (ABCD) là IN. Suy ra: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGnbGaamOtaiaadMeaaiaawkWaaiabg2da9iaaiAdacaaIWaGaeyiS % aalaaa!3D95! \widehat {MNI} = 60^\circ \)

Áp dụng định lí cô sin trong \(\Delta CIN\), ta có

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiaad6 % eacqGH9aqpdaGcaaqaaiaadoeacaWGjbWaaWbaaSqabeaacaaIYaaa % aOGaey4kaSIaam4qaiaad6eadaahaaWcbeqaaiaaikdaaaGccqGHsi % slcaaIYaGaam4qaiaadMeacaGGUaGaam4qaiaad6eacaGGUaGaae4y % aiaab+gacaqGZbGaaGinaiaaiwdacqGHWcaSaSqabaGccqGH9aqpda % GcaaqaamaabmaabaWaaSaaaeaacaaIZaGaamyyamaakaaabaGaaGOm % aaWcbeaaaOqaaiaaisdaaaaacaGLOaGaayzkaaWaaWbaaSqabeaaca % aIYaaaaOGaey4kaSYaaeWaaeaadaWcaaqaaiaadggaaeaacaaIYaaa % aaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaik % dadaWcaaqaaiaaiodacaWGHbWaaOaaaeaacaaIYaaaleqaaaGcbaGa % aGinaaaacaGGUaWaaSaaaeaacaWGHbaabaGaaGOmaaaacaGGUaWaaS % aaaeaadaGcaaqaaiaaikdaaSqabaaakeaacaaIYaaaaaWcbeaakiab % g2da9maalaaabaGaamyyamaakaaabaGaaGynaaWcbeaaaOqaaiaaik % dadaGcaaqaaiaaikdaaSqabaaaaaaa!6568! IN = \sqrt {C{I^2} + C{N^2} - 2CI.CN.{\rm{cos}}45^\circ } = \sqrt {{{\left( {\frac{{3a\sqrt 2 }}{4}} \right)}^2} + {{\left( {\frac{a}{2}} \right)}^2} - 2\frac{{3a\sqrt 2 }}{4}.\frac{a}{2}.\frac{{\sqrt 2 }}{2}} = \frac{{a\sqrt 5 }}{{2\sqrt 2 }}\)

Trong tam giác vuông MIN ta có.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiDaiaacg % gacaGGUbGaaGOnaiaaicdacqGHWcaScqGH9aqpdaWcaaqaaiaad2ea % caWGjbaabaGaamysaiaad6eaaaGaeyO0H4TaamytaiaadMeacqGH9a % qpcaWGjbGaamOtaiaac6cadaGcaaqaaiaaiodaaSqabaGccqGH9aqp % daWcaaqaaiaadggadaGcaaqaaiaaigdacaaI1aaaleqaaaGcbaGaaG % OmamaakaaabaGaaGOmaaWcbeaaaaGccqGH9aqpdaWcaaqaaiaadgga % daGcaaqaaiaaiodacaaIWaaaleqaaaGcbaGaaGinaaaacqGHshI3ca % WGtbGaam4taiabg2da9maalaaabaGaamyyamaakaaabaGaaG4maiaa % icdaaSqabaaakeaacaaIYaaaaaaa!5AA5! \tan 60^\circ = \frac{{MI}}{{IN}} \Rightarrow MI = IN.\sqrt 3 = \frac{{a\sqrt {15} }}{{2\sqrt 2 }} = \frac{{a\sqrt {30} }}{4} \Rightarrow SO = \frac{{a\sqrt {30} }}{2}\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamOqaiaadoeacaGGSaGaamiraiaad2eaaiaawIcacaGLPaaa % cqGH9aqpcaWGKbWaaeWaaeaacaWGcbGaam4qaiaacYcadaqadaqaai % aadofacaWGbbGaamiraaGaayjkaiaawMcaaaGaayjkaiaawMcaaiab % g2da9iaadsgadaqadaqaaiaad6eacaGGSaWaaeWaaeaacaWGtbGaam % yqaiaadseaaiaawIcacaGLPaaaaiaawIcacaGLPaaacqGH9aqpcaaI % YaGaamizamaabmaabaGaam4taiaacYcadaqadaqaaiaadofacaWGbb % GaamiraaGaayjkaiaawMcaaaGaayjkaiaawMcaaiabg2da9iaaikda % caWGKbWaaeWaaeaacaWGpbGaaiilamaabmaabaGaam4uaiaadkeaca % WGdbaacaGLOaGaayzkaaaacaGLOaGaayzkaaaaaa!6222! d\left( {BC,DM} \right) = d\left( {BC,\left( {SAD} \right)} \right) = d\left( {N,\left( {SAD} \right)} \right) = 2d\left( {O,\left( {SAD} \right)} \right) = 2d\left( {O,\left( {SBC} \right)} \right)\)

Kẻ \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x% fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4taiaadw % eacqGHLkIxcaWGtbGaamOtaiabgkDiElaad+eacaWGfbGaeyyPI41a % aeWaaeaacaWGtbGaamOqaiaadoeaaiaawIcacaGLPaaaaaa!448A! OE \bot SN \Rightarrow OE \bot \left( {SBC} \right)\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaam4taiaacYcadaqadaqaaiaadofacaWGcbGaam4qaaGaayjk % aiaawMcaaaGaayjkaiaawMcaaiabg2da9iaad+eacaWGfbaaaa!407E! d\left( {O,\left( {SBC} \right)} \right) = OE\) mà \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaam4taiaadweadaahaaWcbeqaaiaaikdaaaaaaOGaeyyp % a0ZaaSaaaeaacaaIXaaabaGaam4taiaadofadaahaaWcbeqaaiaaik % daaaaaaOGaey4kaSYaaSaaaeaacaaIXaaabaGaam4taiaad6eadaah % aaWcbeqaaiaaikdaaaaaaOGaeyypa0ZaaSaaaeaacaaI0aaabaGaaG % 4maiaaicdacaWGHbWaaWbaaSqabeaacaaIYaaaaaaakiabgUcaRmaa % laaabaGaaGinaaqaaiaadggadaahaaWcbeqaaiaaikdaaaaaaOGaey % ypa0ZaaSaaaeaacaaI2aGaaGOmaaqaaiaaigdacaaI1aGaamyyamaa % CaaaleqabaGaaGOmaaaaaaGccqGHshI3caWGpbGaamyraiabg2da9m % aalaaabaGaamyyamaakaaabaGaaGymaiaaiwdaaSqabaaakeaadaGc % aaqaaiaaiAdacaaIYaaaleqaaaaaaaa!59C6! \frac{1}{{O{E^2}}} = \frac{1}{{O{S^2}}} + \frac{1}{{O{N^2}}} = \frac{4}{{30{a^2}}} + \frac{4}{{{a^2}}} = \frac{{62}}{{15{a^2}}} \Rightarrow OE = \frac{{a\sqrt {15} }}{{\sqrt {62} }}\)

Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamOqaiaadoeacaGGSaGaamiraiaad2eaaiaawIcacaGLPaaa % cqGH9aqpcaaIYaGaam4taiaadweacqGH9aqpdaWcaaqaaiaaikdaca % WGHbWaaOaaaeaacaaIXaGaaGynaaWcbeaaaOqaamaakaaabaGaaGOn % aiaaikdaaSqabaaaaOGaeyypa0ZaaOaaaeaadaWcaaqaaiaaiodaca % aIWaaabaGaaG4maiaaigdaaaaaleqaaOGaamyyaaaa!4AA8! d\left( {BC,DM} \right) = 2OE = \frac{{2a\sqrt {15} }}{{\sqrt {62} }} = \sqrt {\frac{{30}}{{31}}} a\)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iGacYgacaGGVbGaai4zamaaBaaaleaacaaIYaaabeaakiaadgga % caGG7aGaamyEaiabg2da9iGacYgacaGGVbGaai4zamaaBaaaleaaca % aIYaaabeaakiaadkgacaGG7aGaamOEaiabg2da9iGacYgacaGGVbGa % ai4zamaaBaaaleaacaaIYaaabeaakiaadogacaGGUaaaaa!4C2B! x = {\log _2}a;y = {\log _2}b;z = {\log _2}c.\) Vì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiaacY % cacaWGIbGaaiilaiaadogacqGHiiIZdaWadaqaaiaaigdacaGG7aGa % aGPaVlaaikdaaiaawUfacaGLDbaaaaa!4140! a,b,c \in \left[ {1;\,2} \right]\) nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiaacY % cacaaMc8UaaGPaVlaadMhacaGGSaGaaGPaVlaaykW7caWG6bGaeyic % I48aamWaaeaacaaIWaGaai4oaiaaigdaaiaawUfacaGLDbaaaaa!4624! x,\,\,y,\,\,z \in \left[ {0;1} \right]\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaWGqb % Gaeyypa0JaamyyamaaCaaaleqabaGaaG4maaaakiabgUcaRiaadkga % daahaaWcbeqaaiaaiodaaaGccqGHRaWkcaWGJbWaaWbaaSqabeaaca % aIZaaaaOGaeyOeI0IaaG4mamaabmaabaGaciiBaiaac+gacaGGNbWa % aSbaaSqaaiaaikdaaeqaaOGaamyyamaaCaaaleqabaGaamyyaaaaki % abgUcaRiGacYgacaGGVbGaai4zamaaBaaaleaacaaIYaaabeaakiaa % dkgadaahaaWcbeqaaiaadkgaaaGccqGHRaWkciGGSbGaai4BaiaacE % gadaWgaaWcbaGaaGOmaaqabaGccaWGJbWaaWbaaSqabeaacaWGJbaa % aaGccaGLOaGaayzkaaaabaGaaGPaVlaaykW7caaMc8UaaGPaVlaayk % W7cqGH9aqpcaWGHbWaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaamOy % amaaCaaaleqabaGaaG4maaaakiabgUcaRiaadogadaahaaWcbeqaai % aaiodaaaGccqGHsislcaaIZaWaaeWaaeaacaWGHbGaciiBaiaac+ga % caGGNbWaaSbaaSqaaiaaikdaaeqaaOGaamyyaiabgUcaRiaadkgaci % GGSbGaai4BaiaacEgadaWgaaWcbaGaaGOmaaqabaGccaWGIbGaey4k % aSIaam4yaiGacYgacaGGVbGaai4zamaaBaaaleaacaaIYaaabeaaki % aadogaaiaawIcacaGLPaaaaeaacaaMc8UaaGPaVlaaykW7caaMc8Ua % aGPaVlabg2da9iaadggadaahaaWcbeqaaiaaiodaaaGccqGHRaWkca % WGIbWaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaam4yamaaCaaaleqa % baGaaG4maaaakiabgkHiTiaaiodadaqadaqaaiaadggacaWG4bGaey % 4kaSIaamOyaiaadMhacqGHRaWkcaWGJbGaamOEaaGaayjkaiaawMca % aiaac6caaaaa!969B! \begin{array}{l} P = {a^3} + {b^3} + {c^3} - 3\left( {{{\log }_2}{a^a} + {{\log }_2}{b^b} + {{\log }_2}{c^c}} \right)\\ \,\,\,\,\, = {a^3} + {b^3} + {c^3} - 3\left( {a{{\log }_2}a + b{{\log }_2}b + c{{\log }_2}c} \right)\\ \,\,\,\,\, = {a^3} + {b^3} + {c^3} - 3\left( {ax + by + cz} \right). \end{array}\)

Ta chứng minh \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyamaaCa % aaleqabaGaaG4maaaakiabgkHiTiaaiodacaWGHbGaamiEaiabgsMi % JkaadIhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaaIXaGaaiOlaa % aa!4150! {a^3} - 3ax \le {x^3} + 1.\) Thật vậy

Xét hàm số :\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamyyaaGaayjkaiaawMcaaiabg2da9iaadggacqGHsislciGG % SbGaai4BaiaacEgadaWgaaWcbaGaaGOmaaqabaGccaWGHbGaaiilai % aaykW7caWGHbGaeyicI48aamWaaeaacaaIXaGaai4oaiaaykW7caaM % c8UaaGOmaaGaay5waiaaw2faaiabgkDiElqadAgagaqbamaabmaaba % GaamyyaaGaayjkaiaawMcaaiabg2da9iaaigdacqGHsisldaWcaaqa % aiaaigdaaeaacaWGHbGaciiBaiaac6gacaaIYaaaaiabgkDiElqadA % gagaqbamaabmaabaGaamyyaaGaayjkaiaawMcaaiabg2da9iaaicda % cqGHuhY2caWGHbGaeyypa0ZaaSaaaeaacaaIXaaabaGaciiBaiaac6 % gacaaIYaaaaaaa!68AA! f\left( a \right) = a - {\log _2}a,\,a \in \left[ {1;\,\,2} \right] \Rightarrow f'\left( a \right) = 1 - \frac{1}{{a\ln 2}} \Rightarrow f'\left( a \right) = 0 \Leftrightarrow a = \frac{1}{{\ln 2}}\)

Trên đoạn \([1;2]\) ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamyyaaGaayjkaiaawMcaaiabgsMiJkaab2eacaqGHbGaaeiE % amaacmaabaGaamOzamaabmaabaGaaGymaaGaayjkaiaawMcaaiaacY % cacaWGMbWaaeWaaeaacaaIYaaacaGLOaGaayzkaaGaaiilaiaadAga % daqadaqaamaalaaabaGaaGymaaqaaiGacYgacaGGUbGaaGOmaaaaai % aawIcacaGLPaaaaiaawUhacaGL9baacqGH9aqpcaaIXaGaeyO0H4Ta % amyyaiabgkHiTiGacYgacaGGVbGaai4zamaaBaaaleaacaaIYaaabe % aakiaadggacqGHKjYOcaaIXaaaaa!5A8A! f\left( a \right) \le {\rm{Max}}\left\{ {f\left( 1 \right),f\left( 2 \right),f\left( {\frac{1}{{\ln 2}}} \right)} \right\} = 1 \Rightarrow a - {\log _2}a \le 1\)

hay \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabgk % HiTiaadIhacqGHKjYOcaaIXaGaeyi1HSTaamyyaiabgkHiTiaadIha % cqGHsislcaaIXaGaeyizImQaaGimaiaac6caaaa!4529! a - x \le 1 \Leftrightarrow a - x - 1 \le 0.\) Do đó.

Xét: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyamaaCa % aaleqabaGaaG4maaaakiabgkHiTiaaiodacaWGHbGaamiEaiabgkHi % TiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIXaGaeyypa0 % ZaaeWaaeaacaWGHbGaeyOeI0IaamiEaiabgkHiTiaaigdaaiaawIca % caGLPaaadaqadaqaaiaadggadaahaaWcbeqaaiaaikdaaaGccqGHRa % WkcaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGymaiabgUca % RiaadggacqGHRaWkcaWGHbGaamiEaiabgkHiTiaadIhaaiaawIcaca % GLPaaacqGHKjYOcaaIWaaaaa!579E! {a^3} - 3ax - {x^3} - 1 = \left( {a - x - 1} \right)\left( {{a^2} + {x^2} + 1 + a + ax - x} \right) \le 0\)

Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyamaaCa % aaleqabaGaaG4maaaakiabgkHiTiaaiodacaWGHbGaamiEaiabgkHi % TiaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIXaGaeyizIm % QaaGimaaaa!4250! {a^3} - 3ax - {x^3} - 1 \le 0\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % yyamaaCaaaleqabaGaaG4maaaakiabgkHiTiaaiodacaWGHbGaamiE % aiabgsMiJkaadIhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaaIXa % aaaa!42FA! \Leftrightarrow {a^3} - 3ax \le {x^3} + 1\)Tương tự \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % yyamaaCaaaleqabaGaaG4maaaakiabgkHiTiaaiodacaWGHbGaamiE % aiabgsMiJkaadIhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaaIXa % aaaa!42FA! \Leftrightarrow {a^3} - 3ax \le {x^3} + 1\); \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4yamaaCa % aaleqabaGaaG4maaaakiabgkHiTiaaiodacaWGJbGaamOEaiabgsMi % JkaadQhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaaIXaaaaa!40A6! {c^3} - 3cz \le {z^3} + 1\)

Do đó :\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiaayk % W7cqGH9aqpcaWGHbWaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaamOy % amaaCaaaleqabaGaaG4maaaakiabgUcaRiaadogadaahaaWcbeqaai % aaiodaaaGccqGHsislcaaIZaWaaeWaaeaacaWGHbGaamiEaiabgUca % RiaadkgacaWG5bGaey4kaSIaam4yaiaadQhaaiaawIcacaGLPaaaca % aMc8UaeyizImQaamiEamaaCaaaleqabaGaaG4maaaakiabgUcaRiaa % dMhadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaWG6bWaaWbaaSqabe % aacaaIZaaaaOGaey4kaSIaaG4maiabgsMiJkaaigdacqGHRaWkcaaI % ZaGaeyypa0JaaGinaaaa!5DA1! P\, = {a^3} + {b^3} + {c^3} - 3\left( {ax + by + cz} \right)\, \le {x^3} + {y^3} + {z^3} + 3 \le 1 + 3 = 4\)

Đẳng thức xảy ra khi và chỉ khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iaadMhacqGH9aqpcaaIWaGaaiilaiaadQhacqGH9aqpcaaIXaaa % aa!3E25! x = y = 0,z = 1\) và các hoán vị, tức là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabg2 % da9iaadkgacqGH9aqpcaaIXaGaaiilaiaadogacqGH9aqpcaaIYaaa % aa!3DE2! a = b = 1,c = 2\) và các hoán vị. Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabgU % caRiaadkgacqGHRaWkcaWGJbGaeyypa0JaaGinaaaa!3C31! a + b + c = 4\)