Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaiodaaeqaaOWaaSaaaeaacaaIZaaabaGa % aGOmaiaaiwdaaaGaeyypa0JaaGymaiabgkHiTiaaikdaciGGSbGaai % 4BaiaacEgadaWgaaWcbaGaaG4maaqabaGccaaI1aGaeyypa0JaaGym % aiabgkHiTiaaikdacaWGHbaaaa!483A! {\log _3}\frac{3}{{25}} = 1 - 2{\log _3}5 = 1 - 2a\)

Nguyên hàm của hàm số \(f\left( x \right) = 2x + {2^x}\) là: \({x^2} + \frac{{{2^x}}}{{\ln 2}} + C\).

Dưa vào BBT Hàm số đã cho đạt cực tiểu tại x = 2

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maalaaabaGaaGymaaqaaiaaikdaaaGaaiOlamaabmaabaGaaGOm % aiaadggaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGH9a % qpcaaIYaGaamyyamaaCaaaleqabaGaaGOmaaaaaaa!41BA! S = \frac{1}{2}.{\left( {2a} \right)^2} = 2{a^2}\)

Dựa vào hình vẽ, ta thấy f(x)= −2019 < -11; có 2 nghiệm phân biệt.

Suy ra đồ thị hàm sốy=f(x) cắt đường thẳng y = -2019 tại 2 điểm phân biệt.

Ta có 

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqamaabm % aabaGaaGOmaiaacUdacaaIWaaacaGLOaGaayzkaaGaaiilaiaadkea % daqadaqaaiaaicdacaGG7aGaeyOeI0IaaGOmaaGaayjkaiaawMcaai % abgkDiElaadofadaWgaaWcbaGaam4taiaadgeacaWGcbaabeaakiab % g2da9maalaaabaGaaGymaaqaaiaaikdaaaGaaiOlaiaaikdacaGGUa % GaaGOmaiabg2da9iaaikdaaaa!4D90! A\left( {2;0} \right),B\left( {0; - 2} \right) \Rightarrow {S_{OAB}} = \frac{1}{2}.2.2 = 2\)

Mặt cầu (S) có tâm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysamaabm % aabaGaaGymaiaacUdacaaIXaGaai4oaiabgkHiTiaaiodaaiaawIca % caGLPaaacqGHshI3caWGHbGaey4kaSIaamOyaiabgUcaRiaadogacq % GH9aqpcqGHsislcaaIXaaaaa!466C! I\left( {1;1; - 3} \right) \Rightarrow a + b + c = - 1\)

Điều kiện \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgU % caRiaaigdacqGH+aGpcaaIWaGaeyi1HSTaamiEaiabg6da+iabgkHi % Tiaaigdaaaa!4058! x + 1 > 0 \Leftrightarrow x > - 1\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaabm % aabaGaaGOmaiabgkHiTiaadMgaaiaawIcacaGLPaaacqGHRaWkcaaI % XaGaaGOmaiaadMgacqGH9aqpcaaIXaGaeyi1HSTaamOEaiabg2da9m % aalaaabaGaaGymaiabgkHiTiaaigdacaaIYaGaamyAaaqaaiaaikda % cqGHsislcaWGPbaaaiabgsDiBpaaemaabaGaamOEaaGaay5bSlaawI % a7aiabg2da9maalaaabaWaaqWaaeaacaaIXaGaeyOeI0IaaGymaiaa % ikdacaWGPbaacaGLhWUaayjcSdaabaWaaqWaaeaacaaIYaGaeyOeI0 % IaamyAaaGaay5bSlaawIa7aaaacqGHuhY2daabdaqaaiaadQhaaiaa % wEa7caGLiWoacqGH9aqpdaGcaaqaaiaaikdacaaI5aaaleqaaaaa!68C7! z\left( {2 - i} \right) + 12i = 1 \Leftrightarrow z = \frac{{1 - 12i}}{{2 - i}} \Leftrightarrow \left| z \right| = \frac{{\left| {1 - 12i} \right|}}{{\left| {2 - i} \right|}} \Leftrightarrow \left| z \right| = \sqrt {29} \)

Đồ thị hàm số có tiệm cận đứng x = 1, tiệm cận ngang là y = 3 và y = 5.

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaaiodacaWGHbGaey4kaSIaaGOmaiaadkgacqGHRaWkcaWG4bGa % eyOeI0IaaGyoaiabg2da9iaaicdaaeaacqGHsislcaaIZaGaamyyai % abgUcaRiaaiwdacaWGIbGaey4kaSIaaGOmaiaadogacqGHsislcaaI % 5aGaeyypa0JaaGimaaqaaiaaiodacaWGHbGaey4kaSIaamOyaiabgU % caRiaadogacqGH9aqpcaaIWaaaaiaawUhaaiabgsDiBpaaceaaeaqa % beaacaWGHbGaeyypa0JaaGOmaaqaaiaadkgacqGH9aqpcaaI5aaaba % Gaam4yaiabg2da9iabgkHiTiaaigdacaaI1aaaaiaawUhaaiabgkDi % ElaadggacqGHRaWkcaWGIbGaey4kaSIaam4yaiabg2da9iabgkHiTi % aaisdaaaa!6942! \left\{ \begin{array}{l} 3a + 2b + x - 9 = 0\\ - 3a + 5b + 2c - 9 = 0\\ 3a + b + c = 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = 2\\ b = 9\\ c = - 15 \end{array} \right. \Rightarrow a + b + c = - 4\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG4bGaey4kaSYaaSaaaeaacaaI4aaabaGaamiEamaaCaaaleqabaGa % aGOmaaaaaaaakiaawIcacaGLPaaadaahaaWcbeqaaiaaiMdaaaGccq % GH9aqpdaaeWbqaaiaadoeadaqhaaWcbaGaaGyoaaqaaiaadUgaaaGc % caGGUaGaamiEamaaCaaaleqabaGaaGyoaiabgkHiTiaadUgaaaGcda % qadaqaamaalaaabaGaaGioaaqaaiaadIhadaahaaWcbeqaaiaaikda % aaaaaaGccaGLOaGaayzkaaWaaWbaaSqabeaacaWGRbaaaaqaaiaadU % gacqGH9aqpcaaIWaaabaGaaGyoaaqdcqGHris5aOGaeyypa0ZaaabC % aeaacaWGdbWaa0baaSqaaiaaiMdaaeaacaWGRbaaaOGaaiOlaiaaiI % dadaahaaWcbeqaaiaadUgaaaGccaGGUaGaamiEamaaCaaaleqabaGa % aGyoaiabgkHiTiaaiodacaWGRbaaaaqaaiaadUgacqGH9aqpcaaIWa % aabaGaaGyoaaqdcqGHris5aaaa!61A7! {\left( {x + \frac{8}{{{x^2}}}} \right)^9} = \sum\limits_{k = 0}^9 {C_9^k.{x^{9 - k}}{{\left( {\frac{8}{{{x^2}}}} \right)}^k}} = \sum\limits_{k = 0}^9 {C_9^k{{.8}^k}.{x^{9 - 3k}}} \)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGyoaiabgk % HiTiaaiodacaWGRbGaeyypa0JaaGimaiabgsDiBlaadUgacqGH9aqp % caaIZaaaaa!401F! 9- 3k = 0 \Leftrightarrow k = 3\) . Số hạng không chứa x là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qamaaDa % aaleaacaaI5aaabaGaaG4maaaakiaac6cacaaI4aWaaWbaaSqabeaa % caaIZaaaaOGaeyypa0JaaGinaiaaiodacaaIWaGaaGimaiaaiIdaaa % a!3F91! C_9^3{.8^3} = 43008\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmamaaCa % aaleqabaGaamiEamaaCaaameqabaGaaGOmaaaaliabgkHiTiaaigda % aaGccqGH9aqpcaaIZaWaaWbaaSqabeaacaaIYaGaamiEaiabgUcaRi % aaiodaaaGccqGHuhY2caWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOe % I0IaaGymaiabg2da9maabmaabaGaaGOmaiaadIhacqGHRaWkcaaIZa % aacaGLOaGaayzkaaGaciiBaiaac+gacaGGNbWaaSbaaSqaaiaaikda % aeqaaOGaaG4maiabgsDiBlaadIhadaahaaWcbeqaaiaaikdaaaGccq % GHsislcaaIYaGaamiEaiGacYgacaGGVbGaai4zamaaBaaaleaacaaI % YaaabeaakiaaiodacqGHsislcaaIZaGaciiBaiaac+gacaGGNbWaaS % baaSqaaiaaikdaaeqaaOGaaG4maiabgkHiTiaaigdacqGH9aqpcaaI % Waaaaa!652E! {2^{{x^2} - 1}} = {3^{2x + 3}} \Leftrightarrow {x^2} - 1 = \left( {2x + 3} \right){\log _2}3 \Leftrightarrow {x^2} - 2x{\log _2}3 - 3{\log _2}3 - 1 = 0\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBa % aaleaacaaIXaaabeaakiaac6cacaWG4bWaaSbaaSqaaiaaikdaaeqa % aOGaeyypa0JaeyOeI0IaaG4maiGacYgacaGGVbGaai4zamaaBaaale % aacaaIYaaabeaakiaaiodacqGHsislcaaIXaGaeyypa0JaeyOeI0Ia % ciiBaiaac+gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOGaaGynaiaais % daaaa!4A8B! {x_1}.{x_2} = - 3{\log _2}3 - 1 = - {\log _2}54\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGcbGaamyqaiqadgeagaqbaiqadoeagaqbaiaadoeaaeqa % aOGaeyypa0JaamOvaiabgkHiTmaalaaabaGaaGymaaqaaiaaiodaaa % GaamOvaiabg2da9maalaaabaGaaGOmaiaadAfaaeaacaaIZaaaaaaa % !439A! {V_{BAA'C'C}} = V - \frac{1}{3}V = \frac{{2V}}{3}\)

Gọi  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytamaabm % aabaGaamOEamaaBaaaleaacaaIXaaabeaaaOGaayjkaiaawMcaaiab % gkDiElaad2eaaaa!3D6D! M\left( {{z_1}} \right) \Rightarrow M\) thuộc đường tròn \((C_1)\) tâm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysamaaBa % aaleaacaaIXaaabeaakmaabmaabaGaaGymaiaacUdacaaIYaaacaGL % OaGaayzkaaGaaiilaiaadkfadaWgaaWcbaGaaGymaaqabaGccqGH9a % qpcaaIXaaaaa!3FAA! {I_1}\left( {1;2} \right),{R_1} = 1\)

Gọi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOtamaabm % aabaGaamOEamaaBaaaleaacaaIYaaabeaaaOGaayjkaiaawMcaaiab % gkDiElaad6eaaaa!3D70! N\left( {{z_2}} \right) \Rightarrow N\) thuộc đường tròn \((C_2)\) tâm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysamaaBa % aaleaacaaIYaaabeaakmaabmaabaGaaGynaiaacUdacqGHsislcaaI % XaaacaGLOaGaayzkaaGaaiilaiaadkfadaWgaaWcbaGaaGOmaaqaba % GccqGH9aqpcaaIYaaaaa!409D! {I_2}\left( {5; - 1} \right),{R_2} = 2\).

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGjbWaaSbaaSqaaiaaigdaaeqaaOGaamysamaaBaaaleaacaaIYaaa % beaaaOGaay51GaGaeyypa0ZaaeWaaeaacaaI0aGaai4oaiabgkHiTi % aaiodaaiaawIcacaGLPaaacqGHshI3caWGjbWaaSbaaSqaaiaaigda % aeqaaOGaamysamaaBaaaleaacaaIYaaabeaakiabg2da9iaaiwdacq % GH+aGpcaWGsbWaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaamOuamaa % BaaaleaacaaIYaaabeaaaaa!4DEE! \overrightarrow {{I_1}{I_2}} = \left( {4; - 3} \right) \Rightarrow {I_1}{I_2} = 5 > {R_1} + {R_2}\) nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGdbWaaSbaaSqaaiaaigdaaeqaaaGccaGLOaGaayzkaaGaaiilamaa % bmaabaGaam4qamaaBaaaleaacaaIYaaabeaaaOGaayjkaiaawMcaaa % aa!3D28! \left( {{C_1}} \right),\left( {{C_2}} \right)\) không cắt nhau.

Do đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaaBa % aaleaaciGGTbGaaiyAaiaac6gaaeqaaOGaeyypa0Jaamytaiaad6ea % daWgaaWcbaGaciyBaiaacMgacaGGUbaabeaakiabg2da9iaadMeada % WgaaWcbaGaaGymaaqabaGccaWGjbWaaSbaaSqaaiaaikdaaeqaaOGa % eyOeI0IaamOuamaaBaaaleaacaaIXaaabeaakiabgkHiTiaadkfada % WgaaWcbaGaaGOmaaqabaGccqGH9aqpcaaIYaaaaa!4B35! {P_{\min }} = M{N_{\min }} = {I_1}{I_2} - {R_1} - {R_2} = 2\).

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0JaamiEamaaCaaaleqabaGaaG4maaaakiabgkHiTiaad2ga % caWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamiEaiabgkHiTi % aad2gacqGH9aqpcaWG4bWaaeWaaeaacaWG4bWaaWbaaSqabeaacaaI % YaaaaOGaey4kaSIaaGymaaGaayjkaiaawMcaaiabgkHiTiaad2gada % qadaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIXaaa % caGLOaGaayzkaaGaeyypa0ZaaeWaaeaacaWG4bGaeyOeI0IaamyBaa % GaayjkaiaawMcaamaabmaabaGaamiEamaaCaaaleqabaGaaGOmaaaa % kiabgUcaRiaaigdaaiaawIcacaGLPaaacqGHLjYScaaIWaGaeyi1HS % TaamiEaiabgkHiTiaad2gacqGHLjYScaaIWaGaeyi1HSTaamiEaiab % gwMiZkaad2gaaaa!6A59! y' = {x^3} - m{x^2} + x - m = x\left( {{x^2} + 1} \right) - m\left( {{x^2} + 1} \right) = \left( {x - m} \right)\left( {{x^2} + 1} \right) \ge 0 \Leftrightarrow x - m \ge 0 \Leftrightarrow x \ge m\)

Hàm số đã cho đồng biến trên khoảng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aI2aGaai4oaiabgUcaRiabg6HiLcGaayjkaiaawMcaaiabgsDiBlaa % dIhacqGHLjYScaWGTbWaaeWaaeaacqGHaiIicaWG4bGaeyicI48aae % WaaeaacaaI2aGaai4oaiabgUcaRiabg6HiLcGaayjkaiaawMcaaaGa % ayjkaiaawMcaaiabgsDiBlaad2gacqGHKjYOcaaI2aaaaa!5157! \left( {6; + \infty } \right) \Leftrightarrow x \ge m\left( {\forall x \in \left( {6; + \infty } \right)} \right) \Leftrightarrow m \le 6\).

Kết hợp \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaad2gacqGHiiIZcqWIKeIOaeaadaabdaqaaiaad2gaaiaawEa7 % caGLiWoacqGHKjYOcaaIYaGaaGimaiaaikdacaaIWaaaaiaawUhaai % abgsDiBpaaceaaeaqabeaacaWGTbGaeyicI4SaeSijHikabaGaamyB % aiabgIGiopaadmaabaGaeyOeI0IaaGOmaiaaicdacaaIYaGaaGimai % aacUdacaaI2aaacaGLBbGaayzxaaaaaiaawUhaaiabgkDiEdaa!573F! \left\{ \begin{array}{l} m \in Z \\ \left| m \right| \le 2020 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m \in Z\\ m \in Z \left[ { - 2020;6} \right] \end{array} \right. \Rightarrow \) có 2027 giá trị của m

Đường thẳng d đi qua hai điểm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqamaabm % aabaGaeyOeI0IaaGOmaiaacUdacaaIWaaacaGLOaGaayzkaaGaaiil % aiaadkeadaqadaqaaiabgkHiTiaaigdacaGG7aGaaGymaaGaayjkai % aawMcaaiabgkDiElaadsgacaGG6aGaamyEaiabg2da9iaadIhacqGH % RaWkcaaIYaaaaa!4A29! A\left( { - 2;0} \right),B\left( { - 1;1} \right) \Rightarrow d:y = x + 2\).

Phương trình(P)  đỉnh O(0;0) , đi qua B(-1;1) là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iaadIhadaahaaWcbeqaaiaaikdaaaaaaa!39DD! y = {x^2}\).

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaaca % WGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaaGaamizaiaadIhaaSqa % aiabgkHiTiaaiodaaeaacaaIZaaaniabgUIiYdGccqGH9aqpdaWdXb % qaaiaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacaWGKbGaamiE % aaWcbaGaeyOeI0IaaG4maaqaaiabgkHiTiaaigdaa0Gaey4kIipaki % abgUcaRmaapehabaGaamOzamaabmaabaGaamiEaaGaayjkaiaawMca % aiaadsgacaWG4baaleaacqGHsislcaaIXaaabaGaaG4maaqdcqGHRi % I8aOGaeyypa0JaeyOeI0YaaSaaaeaacaaIXaaabaGaaGOmaaaacqGH % RaWkdaWcaaqaaiaaigdaaeaacaaIYaaaaiabgUcaRmaapehabaGaam % iEamaaCaaaleqabaGaaGOmaaaakiaadsgacaWG4baaleaacqGHsisl % caaIXaaabaGaaG4maaqdcqGHRiI8aOGaeyypa0ZaaSaaaeaacaaIYa % GaaGioaaqaaiaaiodaaaaaaa!6A1B! \int\limits_{ - 3}^3 {f\left( x \right)dx} = \int\limits_{ - 3}^{ - 1} {f\left( x \right)dx} + \int\limits_{ - 1}^3 {f\left( x \right)dx} = - \frac{1}{2} + \frac{1}{2} + \int\limits_{ - 1}^3 {{x^2}dx} = \frac{{28}}{3}\).

Tập hợp các điểm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytamaaBa % aaleaacaaIXaaabeaakiaacYcacaWGnbWaaSbaaSqaaiaaikdaaeqa % aaaa!3A20! {M_1},{M_2}\) biểu diễn số phức \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaBa % aaleaacaaIXaaabeaakiaacYcacaWG6bWaaSbaaSqaaiaaikdaaeqa % aaaa!3A7A! {z_1},{z_2}\) là các đường tròn đồng tâm I(-2;-3) , bán kính lần lượt là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuamaaBa % aaleaacaaIXaaabeaakiabg2da9iaaiodacaGGSaGaamOuamaaBaaa % leaacaaIYaaabeaakiabg2da9maalaaabaGaaG4maaqaaiaaiwdaaa % aaaa!3E89! {R_1} = 3,{R_2} = \frac{3}{5}\).

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG6bWaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaaGOmaiabgUcaRiaa % iodacaWGPbaabaGaamOEamaaBaaaleaacaaIYaaabeaakiabgUcaRi % aaikdacqGHRaWkcaaIZaGaamyAaaaacqGH9aqpcaWG4bGaey4kaSIa % amyEaiaadMgacqGHshI3daabdaqaamaalaaabaGaamOEamaaBaaale % aacaaIXaaabeaakiabgUcaRiaaikdacqGHRaWkcaaIZaGaamyAaaqa % aiaadQhadaWgaaWcbaGaaGOmaaqabaGccqGHRaWkcaaIYaGaey4kaS % IaaG4maiaadMgaaaaacaGLhWUaayjcSdGaeyypa0ZaaqWaaeaacaWG % 4bGaey4kaSIaamyEaiaadMgaaiaawEa7caGLiWoacqGHuhY2daabda % qaamaalaaabaGaamOEamaaBaaaleaacaaIXaaabeaakiabgUcaRiaa % ikdacqGHRaWkcaaIZaGaamyAaaqaaiaadQhadaWgaaWcbaGaaGOmaa % qabaGccqGHRaWkcaaIYaGaey4kaSIaaG4maiaadMgaaaaacaGLhWUa % ayjcSdGaeyypa0ZaaOaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaO % Gaey4kaSIaamyEamaaCaaaleqabaGaaGOmaaaaaeqaaOGaeyi1HS9a % aOaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamyEam % aaCaaaleqabaGaaGOmaaaaaeqaaOGaeyypa0JaaGynaaaa!8156! \frac{{{z_1} + 2 + 3i}}{{{z_2} + 2 + 3i}} = x + yi \Rightarrow \left| {\frac{{{z_1} + 2 + 3i}}{{{z_2} + 2 + 3i}}} \right| = \left| {x + yi} \right| \Leftrightarrow \left| {\frac{{{z_1} + 2 + 3i}}{{{z_2} + 2 + 3i}}} \right| = \sqrt {{x^2} + {y^2}} \Leftrightarrow \sqrt {{x^2} + {y^2}} = 5\)

Do \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEamaaCa % aaleqabaGaaGOmaaaakiabgwMiZkaaicdacqGHshI3caWG4bWaaWba % aSqabeaacaaIYaaaaOGaeyizImQaaGOmaiaaiwdacqGHshI3caWG4b % GaeyizImQaaGynaaaa!47AF! {y^2} \ge 0 \Rightarrow {x^2} \le 25 \Rightarrow x \le 5\). Dấu bằng xảy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WG6bWaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaaGOmaiabgUcaRiaa % iodacaWGPbaabaGaamOEamaaBaaaleaacaaIYaaabeaakiabgUcaRi % aaikdacqGHRaWkcaaIZaGaamyAaaaacqGH9aqpcaaI1aaaaa!43FF! \frac{{{z_1} + 2 + 3i}}{{{z_2} + 2 + 3i}} = 5\).

Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBamaaBa % aaleaacaaIWaaabeaakiabg2da9iaaiwdacqGHuhY2daWhcaqaaiaa % dMeacaWGnbWaaSbaaSqaaiaaigdaaeqaaaGccaGLxdcaaaa!403B! {m_0} = 5 \Leftrightarrow \overrightarrow {I{M_1}} \) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGjbGaamytamaaBaaaleaacaaIYaaabeaaaOGaay51Gaaaaa!3A39! \overrightarrow {I{M_2}} \) là hai vecto cùng hướng.

Gọi A,B,C lần lượt là tâm của 3 đường tròn ở 3 góc ngoài cùng.

Khi đó tam giác ABC là tam giác đều cạnh \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOCaiabgU % caRiaaiodacaGGUaWaaeWaaeaacaaIYaGaamOCaaGaayjkaiaawMca % aiabgUcaRiaadkhacqGH9aqpcaaI4aGaamOCaiabg2da9iaaiIdaaa % a!42D7! r + 3.\left( {2r} \right) + r = 8r = 8\).

Chiều cao CK của tam giác là :\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qaiaadU % eacqGH9aqpcaaI4aGaaiOlamaalaaabaWaaOaaaeaacaaIZaaaleqa % aaGcbaGaaGOmaaaacqGH9aqpcaaI0aWaaOaaaeaacaaIZaaaleqaaa % aa!3E4F! CK = 8.\frac{{\sqrt 3 }}{2} = 4\sqrt 3 \)

Chiều cao của đống rơm là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiaadI % eacqGH9aqpcaWGYbGaey4kaSIaam4qaiaadUeacqGHRaWkcaWGYbGa % eyypa0JaaGOmaiaadkhacqGHRaWkcaaI0aWaaOaaaeaacaaIZaaale % qaaOGaeyypa0JaaGOmaiabgUcaRiaaisdadaGcaaqaaiaaiodaaSqa % baaaaa!475D! SH = r + CK + r = 2r + 4\sqrt 3 = 2 + 4\sqrt 3 \)

.

Đặt t = 2x - 1 thì với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaadmaabaGaaGimaiaacUdacaaIXaaacaGLBbGaayzxaaGaeyO0 % H4TaamiDaiabgIGiopaadmaabaGaeyOeI0IaaGymaiaacUdacaaIXa % aacaGLBbGaayzxaaaaaa!4688! x \in \left[ {0;1} \right] \Rightarrow t \in \left[ { - 1;1} \right]\) và với mỗi giá trị của t có một giá trị của x .

Phương trình trở thành \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiDaaGaayjkaiaawMcaaiabg2da9maalaaabaGaamyBaiab % gkHiTiaaikdaaeaacaaIZaaaaaaa!3DCE! f\left( t \right) = \frac{{m - 2}}{3}\) có 3 nghiệm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabgI % GiopaadmaabaGaeyOeI0IaaGymaiaacUdacaaIXaaacaGLBbGaayzx % aaGaeyi1HS9aaSaaaeaacaWGTbGaeyOeI0IaaGOmaaqaaiaaiodaaa % Gaeyypa0JaaGymaiabgsDiBlaad2gacqGH9aqpcaaI1aaaaa!4A1C! t \in \left[ { - 1;1} \right] \Leftrightarrow \frac{{m - 2}}{3} = 1 \Leftrightarrow m = 5\).

Vậy  m = 5.

Bất phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % yBaiabgYda8iaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH % sislcaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiaadI % hacqGH9aqpcaWGNbWaaeWaaeaacaWG4baacaGLOaGaayzkaaaaaa!47A6! \Leftrightarrow m < f\left( x \right) - {x^2} + 2x = g\left( x \right)\) đúng với mọi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaabmaabaGaaGymaiaacUdacaaIYaaacaGLOaGaayzkaaGaaGPa % VlaaykW7caaMc8UaaiikaiaacQcacaGGPaaaaa!42DB! x \in \left( {1;2} \right)\,\,\,(*)\).

Xét \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadAgadaqadaqaaiaa % dIhaaiaawIcacaGLPaaacqGHsislcaWG4bWaaWbaaSqabeaacaaIYa % aaaOGaey4kaSIaaGOmaiaadIhaaaa!4354! g\left( x \right) = f\left( x \right) - {x^2} + 2x\) với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaabmaabaGaaGymaiaacUdacaaIYaaacaGLOaGaayzkaaaaaa!3C33! x \in \left( {1;2} \right)\) ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4zayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JabmOzayaafaWa % aeWaaeaacaWG4baacaGLOaGaayzkaaGaeyOeI0IaaGOmaiaadIhacq % GHRaWkcaaIYaGaeyypa0JabmOzayaafaWaaeWaaeaacaWG4baacaGL % OaGaayzkaaGaeyOeI0IaaGOmamaabmaabaGaamiEaiabgkHiTiaaig % daaiaawIcacaGLPaaaaaa!4C92! g'\left( x \right) = f'\left( x \right) - 2x + 2 = f'\left( x \right) - 2\left( {x - 1} \right)\).

Với \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiopaabmaabaGaaGymaiaacUdacaaIYaaacaGLOaGaayzkaaaaaa!3C33! x \in \left( {1;2} \right)\) thì f'(x) < 0 và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyOeI0IaaG % OmamaabmaabaGaamiEaiabgkHiTiaaigdaaiaawIcacaGLPaaacqGH % 8aapcaaIWaGaeyO0H4Tabm4zayaafaWaaeWaaeaacaWG4baacaGLOa % GaayzkaaGaeyipaWJaaGimamaabmaabaGaeyiaIiIaamiEaiabgIGi % opaabmaabaGaaGymaiaacUdacaaIYaaacaGLOaGaayzkaaaacaGLOa % Gaayzkaaaaaa!4DBA! - 2\left( {x - 1} \right) < 0 \Rightarrow g'\left( x \right) < 0\left( {\forall x \in \left( {1;2} \right)} \right)\).

Do đó hàm số g(x) nghịch biến trên khoảng (1;2).

Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % GGQaaacaGLOaGaayzkaaGaeyi1HSTaamyBaiabgsMiJkaadEgadaqa % daqaaiaaikdaaiaawIcacaGLPaaacqGHuhY2caWGTbGaeyizImQaam % OzamaabmaabaGaaGOmaaGaayjkaiaawMcaaaaa!4891! \left( * \right) \Leftrightarrow m \le g\left( 2 \right) \Leftrightarrow m \le f\left( 2 \right)\).

TH1 : Phương trình có nghiệm thực z thỏa  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG6baacaGLhWUaayjcSdGaeyypa0ZaaOaaaeaacaaIZaaaleqaaOGa % eyO0H4TaamyBaiabg2da9iaaikdacaGGUaaaaa!41C0! \left| z \right| = \sqrt 3 \Rightarrow m = 2.\)

TH2 : Phương trình không có nghiệm thực, khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyamaaCa % aaleqabaGaaGOmaaaakiabgkHiTiaaikdacaWGHbGaeyOpa4JaaGim % aaaa!3C1D! {a^2} - 2a > 0\).

Do a là số thực nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEamaaBa % aaleaacaaIXaGaaiilaiaaikdaaeqaaOGaeyypa0ZaaSaaaeaacqGH % sisldaGcaaqaaiaaiodaaSqabaGccqGHXcqScaWGPbWaaOaaaeaacq % qHuoaraSqabaaakeaacaaIYaaaaaaa!4157! {z_{1,2}} = \frac{{ - \sqrt 3 \pm i\sqrt \Delta }}{2}\) là hai số phức liên hợp của nhau

Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG6bWaaSbaaSqaaiaaigdaaeqaaaGccaGLhWUaayjcSdGaeyypa0Za % aqWaaeaacaWG6bWaaSbaaSqaaiaaikdaaeqaaaGccaGLhWUaayjcSd % aaaa!411E! \left| {{z_1}} \right| = \left| {{z_2}} \right|\), mặt khác \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG6bWaaSbaaSqaaiaaigdaaeqaaOGaaiOlaiaadQhadaWgaaWcbaGa % aGOmaaqabaaakiaawEa7caGLiWoacqGH9aqpdaabdaqaaiaadQhada % WgaaWcbaGaaGymaaqabaaakiaawEa7caGLiWoacaGGUaWaaqWaaeaa % caWG6bWaaSbaaSqaaiaaikdaaeqaaaGccaGLhWUaayjcSdGaeyypa0 % JaamyyamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaikdacaWGHbGa % eyi1HS9aaqWaaeaacaWG6bWaaSbaaSqaaiaaigdaaeqaaaGccaGLhW % UaayjcSdGaeyypa0ZaaqWaaeaacaWG6bWaaSbaaSqaaiaaikdaaeqa % aaGccaGLhWUaayjcSdGaeyypa0ZaaOaaaeaacaWGHbWaaWbaaSqabe % aacaaIYaaaaOGaeyOeI0IaaGOmaiaadggaaSqabaaaaa!6203! \left| {{z_1}.{z_2}} \right| = \left| {{z_1}} \right|.\left| {{z_2}} \right| = {a^2} - 2a \Leftrightarrow \left| {{z_1}} \right| = \left| {{z_2}} \right| = \sqrt {{a^2} - 2a} \).

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yyamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaikdacaWGHbGaeyyp % a0JaaG4maiabgsDiBpaadeaaeaqabeaacaWGHbGaeyypa0JaeyOeI0 % IaaGymaaqaaiaadggacqGH9aqpcaaIZaaaaiaawUfaaaaa!4816! \Rightarrow {a^2} - 2a = 3 \Leftrightarrow \left[ \begin{array}{l} a = - 1\\ a = 3 \end{array} \right.\)

Loại a = -1 . Do đó có 2 giá trị tương đương của a thỏa mãn yêu cầu bài toán

Dựa vào hình vẽ ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4bWaaSbaaSqaaiaadgeaaeqaaaGccaGLOaGaayzk % aaGaeyypa0JaaGimaiaacYcaceWGMbGbauaadaqadaqaaiaadIhada % WgaaWcbaGaamOqaaqabaaakiaawIcacaGLPaaacqGH8aapcaaIWaGa % aiilaiqadAgagaqbamaabmaabaGaamiEamaaBaaaleaacaWGdbaabe % aaaOGaayjkaiaawMcaaiabg6da+iaaicdaaaa!4A01! f'\left( {{x_A}} \right) = 0,f'\left( {{x_B}} \right) < 0,f'\left( {{x_C}} \right) > 0\)

Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4bWaaSbaaSqaaiaadkeaaeqaaaGccaGLOaGaayzk % aaGaeyipaWJabmOzayaafaWaaeWaaeaacaWG4bWaaSbaaSqaaiaadg % eaaeqaaaGccaGLOaGaayzkaaGaeyipaWJabmOzayaafaWaaeWaaeaa % caWG4bWaaSbaaSqaaiaadoeaaeqaaaGccaGLOaGaayzkaaaaaa!4569! f'\left( {{x_B}} \right) < f'\left( {{x_A}} \right) < f'\left( {{x_C}} \right)\)

Khối nón (chiều cao h) nội tiếp khối cầu (bán kính R) có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaaciGGTbGaaiyyaiaacIhaaeqaaOGaeyi1HS9aauIhaeaacaWG % ObGaeyypa0ZaaSaaaeaacaaI0aGaamOuaaqaaiaaiodaaaaaaaaa!40D0! {V_{\max }} \Leftrightarrow h = \frac{4R}{3}\).

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabg2 % da9maalaaabaGaamyqaiaadkeaaeaacaaIYaaaaiabg2da9iaaioda % cqGHshI3caWGObGaeyypa0JaaGinaiabgkDiEpaaFiaabaGaamyqai % aadIeaaiaawEniaiabg2da9maalaaabaGaaGOmaaqaaiaaiodaaaWa % a8HaaeaacaWGbbGaamOqaaGaay51GaGaeyO0H4Taamisamaabmaaba % WaaSaaaeaacaaIXaGaaGinaaqaaiaaiodaaaGaai4oamaalaaabaGa % aGymaiaaigdaaeaacaaIZaaaaiaacUdadaWcaaqaaiaaigdacaaIZa % aabaGaaG4maaaaaiaawIcacaGLPaaaaaa!596D! R = \frac{{AB}}{2} = 3 \Rightarrow h = 4 \Rightarrow \overrightarrow {AH} = \frac{2}{3}\overrightarrow {AB} \Rightarrow H\left( {\frac{{14}}{3};\frac{{11}}{3};\frac{{13}}{3}} \right)\)

Vì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadk % eacqGHLkIxcaWGTbGaamiCamaabmaabaGaamiuaaGaayjkaiaawMca % aiabgkDiEpaaFiaabaGaamOBaaGaay51GaWaaSbaaSqaamaabmaaba % GaamiuaaGaayjkaiaawMcaaaqabaGccqGH9aqpdaWhcaqaaiaadgea % caWGcbaacaGLxdcacaGG7aGaamisaaaa!4AE1! AB \bot mp\left( P \right) \Rightarrow {\overrightarrow n _{\left( P \right)}} = \overrightarrow {AB} ;H\) thuộc mặt phẳng (P).

Phương trình mặt phẳng (P) là:

\(2\left( {x - \frac{{14}}{3}} \right) + 2\left( {y - \frac{{11}}{3}} \right) + z - \frac{{13}}{3} = 0 \Leftrightarrow 2x + 2y + z - 21 = 0\)

Vậy b = 2, c = 1 , d = -21. Nên S = 2+1+(-21) = -18

Đặt t = 3cosx + +2  mà \(m \in (\frac{-\pi}{2};\frac{\pi}{2})\)\(\Rightarrow cosx \in (0;1] \Rightarrow t \in (2;5]\)

Do đó phương trình trở thành : f(t) = m

Yêu cầu bài toán trở thành f(t) = m  có nghiệm thuộc {2;5]khi và chỉ khi \(1 \le m < 3\)

Giả thiết trở thành

 \(\begin{array}{l} \frac{1}{{\sqrt 2 }}\left[ {2xf'(x) + f(x)} \right] = 6\sqrt x \\ \Leftrightarrow 2\sqrt x .f'(x) + \frac{1}{{\sqrt x }}.f(x) = 6\sqrt x \end{array}\) \( \Leftrightarrow 2\sqrt x .f'\left( x \right) + {\left( {2\sqrt x } \right)^\prime }.f\left( x \right) = 6\sqrt x \\ \Leftrightarrow {\left[ {2\sqrt x .f\left( x \right)} \right]^\prime } = 6\sqrt x \Leftrightarrow 2\sqrt x .f\left( x \right) = \int {6\sqrt x dx} \)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaaG % OmamaakaaabaGaamiEaaWcbeaakiaac6cacaWGMbWaaeWaaeaacaWG % 4baacaGLOaGaayzkaaGaeyypa0JaaGinaiaadIhadaGcaaqaaiaadI % haaSqabaGccqGHRaWkcaWGdbaaaa!43DD! \Leftrightarrow 2\sqrt x .f\left( x \right) = 4x\sqrt x + C\) mà \(f\left( 1 \right) = 5 \Rightarrow 2f\left( 1 \right) = 4 + C \Rightarrow C = 6\)

\(\begin{array}{l} f\left( x \right) = \frac{{4x\sqrt x + 6}}{{2\sqrt x }} = 2x + \frac{3}{{\sqrt x }}\\ \Rightarrow \int\limits_4^9 {f\left( x \right)dx} = \int\limits_4^9 {\left( {2x + \frac{3}{{\sqrt x }}} \right)dx = 71} \end{array}\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaaGinaiaadgga % caWG4bGaaiOlamaabmaabaGaamiEaiabgkHiTiaaigdaaiaawIcaca % GLPaaadaqadaqaaiaadIhacqGHsislcaaIYaaacaGLOaGaayzkaaGa % eyypa0JaaGinaiaadggacaWG4bWaaWbaaSqabeaacaaIZaaaaOGaey % OeI0IaaGymaiaaikdacaWGHbGaamiEamaaCaaaleqabaGaaGOmaaaa % kiabgUcaRiaaiIdacaWGHbGaamiEaaaa!5382! f'\left( x \right) = 4ax.\left( {x - 1} \right)\left( {x - 2} \right) = 4a{x^3} - 12a{x^2} + 8ax\)

Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maapeaabaGabmOzayaa % faWaaeWaaeaacaWG4baacaGLOaGaayzkaaGaamizaiaadIhaaSqabe % qaniabgUIiYdGccqGH9aqpcaWGHbGaamiEamaaCaaaleqabaGaaGin % aaaakiabgkHiTiaaisdacaWGHbGaamiEamaaCaaaleqabaGaaG4maa % aakiabgUcaRiaaisdacaWGHbGaamiEamaaCaaaleqabaGaaGOmaaaa % kiabgUcaRiaadwgacqGHshI3caWGIbGaeyypa0JaeyOeI0IaaGinai % aadggacaGG7aGaam4yaiabg2da9iaaisdacaWGHbGaai4oaiaadsga % cqGH9aqpcaaIWaaaaa!5F08! f\left( x \right) = \int {f'\left( x \right)dx} = a{x^4} - 4a{x^3} + 4a{x^2} + e \Rightarrow b = - 4a;c = 4a;d = 0\)

Vậy  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaikdacaWGHbGaey4k % aSIaamOyaiabgUcaRiaadogacqGHRaWkcaWGKbGaey4kaSIaamyzai % abg2da9iaaikdacaWGHbGaey4kaSIaamyzaiabgsDiBlaadggacaWG % 4bWaaWbaaSqabeaacaaI0aaaaOGaeyOeI0IaaGinaiaadggacaWG4b % WaaWbaaSqabeaacaaIZaaaaOGaey4kaSIaaGinaiaadggacaWG4bWa % aWbaaSqabeaacaaIYaaaaOGaey4kaSIaamyzaiabg2da9iaaikdaca % WGHbGaey4kaSIaamyzaaaa!5C16! f\left( x \right) = 2a + b + c + d + e = 2a + e \Leftrightarrow a{x^4} - 4a{x^3} + 4a{x^2} + e = 2a + e\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % iEamaaCaaaleqabaGaaGinaaaakiabgkHiTiaaisdacaWG4bWaaWba % aSqabeaacaaIZaaaaOGaey4kaSIaaGinaiaadIhadaahaaWcbeqaai % aaikdaaaGccqGHsislcaaIYaGaeyypa0JaaGimaiabgsDiBpaabmaa % baGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaikdacaWG4b % aacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0YaaeWa % aeaadaGcaaqaaiaaikdaaSqabaaakiaawIcacaGLPaaadaahaaWcbe % qaaiaaikdaaaGccqGH9aqpcaaIWaGaeyi1HS9aamqaaqaabeqaaiaa % dIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaIYaGaamiEaiabg2 % da9maakaaabaGaaGOmaaWcbeaaaOqaaiaadIhadaahaaWcbeqaaiaa % ikdaaaGccqGHsislcaaIYaGaamiEaiabg2da9iabgkHiTmaakaaaba % GaaGOmaaWcbeaaaaGccaGLBbaaaaa!6593! \Leftrightarrow {x^4} - 4{x^3} + 4{x^2} - 2 = 0 \Leftrightarrow {\left( {{x^2} - 2x} \right)^2} - {\left( {\sqrt 2 } \right)^2} = 0 \Leftrightarrow \left[ \begin{array}{l} {x^2} - 2x = \sqrt 2 \\ {x^2} - 2x = - \sqrt 2 \end{array} \right.\)

Do đó phương trình đã cho có hai nghiệm phân biệt.

Hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaikdacaaIWaGaaGym % aiaaiMdadaahaaWcbeqaaiaadIhaaaGccqGHsislcaaIYaGaaGimai % aaigdacaaI5aWaaWbaaSqabeaacqGHsislcaWG4baaaaaa!448A! f\left( x \right) = {2019^x} - {2019^{ - x}}\) xác định với mọi x thuộc R.

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaeyOeI0IaamiEaaGaayjkaiaawMcaaiabg2da9iaaikdacaaI % WaGaaGymaiaaiMdadaahaaWcbeqaaiabgkHiTiaadIhaaaGccqGHsi % slcaaIYaGaaGimaiaaigdacaaI5aWaaWbaaSqabeaacaWG4baaaOGa % eyypa0JaeyOeI0YaaeWaaeaacaaIYaGaaGimaiaaigdacaaI5aWaaW % baaSqabeaacaWG4baaaOGaeyOeI0IaaGOmaiaaicdacaaIXaGaaGyo % amaaCaaaleqabaGaeyOeI0IaamiEaaaaaOGaayjkaiaawMcaaiabg2 % da9iabgkHiTiaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH % shI3caWGMbWaaeWaaeaacaWG4baacaGLOaGaayzkaaaaaa!5E59! f\left( { - x} \right) = {2019^{ - x}} - {2019^x} = - \left( {{{2019}^x} - {{2019}^{ - x}}} \right) = - f\left( x \right) \Rightarrow f\left( x \right)\) là hàm số lẻ

Mặt khác \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaaGOmaiaaicda % caaIXaGaaGyoamaaCaaaleqabaGaamiEaaaakiGacYgacaGGUbGaaG % OmaiaaicdacaaIXaGaaGyoaiabgUcaRiaaikdacaaIWaGaaGymaiaa % iMdadaahaaWcbeqaaiabgkHiTiaadIhaaaGcciGGSbGaaiOBaiaaik % dacaaIWaGaaGymaiaaiMdacqGH+aGpcaaIWaGaeyiaIiIaamiEaiab % gIGiolabl2riHkabgkDiElaadAgadaqadaqaaiaadIhaaiaawIcaca % GLPaaaaaa!5A96! f'\left( x \right) = {2019^x}\ln 2019 + {2019^{ - x}}\ln 2019 > 0;\forall x \in R \Rightarrow f\left( x \right)\) đồng biến trên R.

Do đó BPT :\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamyBaaGaayjkaiaawMcaaiabgUcaRiaadAgadaqadaqaaiaa % ikdacaWGTbGaey4kaSIaaGOmaiaaicdacaaIXaGaaGyoaaGaayjkai % aawMcaaiabgYda8iaaicdacqGHuhY2caWGMbWaaeWaaeaacaaIYaGa % amyBaiabgUcaRiaaikdacaaIWaGaaGymaiaaiMdaaiaawIcacaGLPa % aacqGH8aapcqGHsislcaWGMbWaaeWaaeaacaWGTbaacaGLOaGaayzk % aaGaeyi1HSTaamOzamaabmaabaGaaGOmaiaad2gacqGHRaWkcaaIYa % GaaGimaiaaigdacaaI5aaacaGLOaGaayzkaaGaeyipaWJaamOzamaa % bmaabaGaeyOeI0IaamyBaaGaayjkaiaawMcaaaaa!6347! f\left( m \right) + f\left( {2m + 2019} \right) < 0 \Leftrightarrow f\left( {2m + 2019} \right) < - f\left( m \right) \Leftrightarrow f\left( {2m + 2019} \right) < f\left( { - m} \right)\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaaG % Omaiaad2gacqGHRaWkcaaIYaGaaGimaiaaigdacaaI5aGaeyipaWJa % eyOeI0IaamyBaiabgsDiBlaad2gacqGH8aapcqGHsislcaaI2aGaaG % 4naiaaiodaaaa!4833! \Leftrightarrow 2m + 2019 < - m \Leftrightarrow m < - 673\)

Mặt cầu có tâm I(1;-2;3) và bán kính \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOuaiabg2 % da9iaaiodadaGcaaqaaiaaiodaaSqabaaaaa!3965! R = 3\sqrt 3 \).

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiaadI % eacqGH9aqpcaWGObGaeyO0H4TaamisaiaadgeadaahaaWcbeqaaiaa % ikdaaaGccqGHsislcaWGObWaaWbaaSqabeaacaaIYaaaaOGaeyypa0 % JaaGOmaiaaiEdacqGHsislcaWGObWaaWbaaSqabeaacaaIYaaaaaaa % !4677! IH = h \Rightarrow H{A^2} - {h^2} = 27 - {h^2}\)

Thể tích khối nón đỉnh I đáy là đường tròn ( C) là : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvaiabg2 % da9maalaaabaGaaGymaaqaaiaaiodaaaGaeqiWdaNaamisaiaadgea % daahaaWcbeqaaiaaikdaaaGccaGGUaGaamiAaiabg2da9maalaaaba % GaaGymaaqaaiaaiodaaaGaeqiWda3aaeWaaeaacaaIYaGaaG4naiab % gkHiTiaadIgadaahaaWcbeqaaiaaikdaaaaakiaawIcacaGLPaaaca % WGObGaeyypa0ZaaSaaaeaacaaIXaaabaGaaG4maaaacqaHapaCdaqa % daqaaiaaikdacaaI3aGaamiAaiabgkHiTiaadIgadaahaaWcbeqaai % aaiodaaaaakiaawIcacaGLPaaaaaa!5555! V = \frac{1}{3}\pi H{A^2}.h = \frac{1}{3}\pi \left( {27 - {h^2}} \right)h = \frac{1}{3}\pi \left( {27h - {h^3}} \right)\)

Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOvayaafa % Gaeyypa0ZaaSaaaeaacaaIXaaabaGaaG4maaaacqaHapaCdaqadaqa % aiaaikdacaaI3aGaeyOeI0IaaG4maiaadIgadaahaaWcbeqaaiaaik % daaaaakiaawIcacaGLPaaacqGH9aqpcaaIWaGaeyi1HSTaamiAaiab % g2da9iaaiodaaaa!4881! V' = \frac{1}{3}\pi \left( {27 - 3{h^2}} \right) = 0 \Leftrightarrow h = 3\)

Từ đó suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaaciGGTbGaaiyyaiaacIhaaeqaaOGaeyi1HSTaamiAaiabg2da % 9iaaiodacqGHshI3caWGKbWaaeWaaeaacaWGjbGaai4oamaabmaaba % GaeqySdegacaGLOaGaayzkaaaacaGLOaGaayzkaaGaeyypa0JaaG4m % aaaa!4A2B! {V_{\max }} \Leftrightarrow h = 3 \Rightarrow d\left( {I;\left( \alpha \right)} \right) = 3\).

Mặt phẳng \((\alpha)\) qua \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqamaabm % aabaGaaGimaiaacUdacaaIWaGaai4oaiabgkHiTiaaisdaaiaawIca % caGLPaaacaGGSaGaamOqamaabmaabaGaaGOmaiaacUdacaaIWaGaai % 4oaiaaicdaaiaawIcacaGLPaaaaaa!438D! A\left( {0;0; - 4} \right),B\left( {2;0;0} \right)\) và cách I một khoảng là 3.

Ta có : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WGUbWaaSbaaSqaamaabmaabaGaamiuaaGaayjkaiaawMcaaaqabaaa % kiaawEniaiabg2da9maabmaabaGaamyyaiaacUdacaWGIbGaai4oai % abgkHiTiaaigdaaiaawIcacaGLPaaacaGGSaWaa8HaaeaacaWGbbGa % amOqaaGaay51GaWaaeWaaeaacaaIYaGaai4oaiaaicdacaGG7aGaaG % inaaGaayjkaiaawMcaaiabgkDiEpaaFiaabaGaamOBamaaBaaaleaa % daqadaqaaiaadcfaaiaawIcacaGLPaaaaeqaaaGccaGLxdcacaGGUa % Waa8HaaeaacaWGbbGaamOqaaGaay51GaGaeyypa0JaaGOmaiaadgga % cqGHsislcaaI0aGaeyypa0JaaGimaiabgsDiBlaadggacqGH9aqpca % aIYaaaaa!627E! \overrightarrow {{n_{\left( P \right)}}} = \left( {a;b; - 1} \right),\overrightarrow {AB} \left( {2;0;4} \right) \Rightarrow \overrightarrow {{n_{\left( P \right)}}} .\overrightarrow {AB} = 2a - 4 = 0 \Leftrightarrow a = 2\)

Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGqbaacaGLOaGaayzkaaGaaiOoaiaaikdacaWG4bGaey4kaSIaamOy % aiaadMhacqGHsislcaWG6bGaeyOeI0IaaGinaiabg2da9iaaicdaaa % a!42E6! \left( P \right):2x + by - z - 4 = 0\). Mặt khác \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamysaiaacUdadaqadaqaaiabeg7aHbGaayjkaiaawMcaaaGa % ayjkaiaawMcaaiabg2da9maalaaabaWaaqWaaeaacaaIYaGaeyOeI0 % IaaGOmaiaadkgacqGHsislcaaI3aaacaGLhWUaayjcSdaabaWaaOaa % aeaacaaIYaWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamOyamaaCa % aaleqabaGaaGOmaaaakiabgUcaRiaaigdadaahaaWcbeqaaiaaikda % aaaabeaaaaGccqGH9aqpcaaIZaaaaa!4F1A! d\left( {I;\left( \alpha \right)} \right) = \frac{{\left| {2 - 2b - 7} \right|}}{{\sqrt {{2^2} + {b^2} + {1^2}} }} = 3\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aae % WaaeaacaaIYaGaamOyaiabgUcaRiaaiwdaaiaawIcacaGLPaaadaah % aaWcbeqaaiaaikdaaaGccqGH9aqpcaaI5aWaaeWaaeaacaaI1aGaey % 4kaSIaamOyamaaCaaaleqabaGaaGOmaaaaaOGaayjkaiaawMcaaiab % gsDiBlaaiwdacaWGIbWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaG % OmaiaaicdacaWGIbGaey4kaSIaaGOmaiaaicdacqGH9aqpcaaIWaGa % eyi1HSTaamOyaiabg2da9iaaikdacqGHshI3caWGHbGaey4kaSIaam % OyaiabgUcaRiaadsgacqGH9aqpcaaIYaGaey4kaSIaaGOmaiabgkHi % TiaaisdacqGH9aqpcaaIWaaaaa!63DA! \Leftrightarrow {\left( {2b + 5} \right)^2} = 9\left( {5 + {b^2}} \right) \Leftrightarrow 5{b^2} - 20b + 20 = 0 \Leftrightarrow b = 2 \Rightarrow a + b + d = 2 + 2 - 4 = 0\)

. 

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaada % WcaaqaamaabmaabaGaaGymaiaaikdacqGHsislcaaI1aGaamyAaaGa % ayjkaiaawMcaaiaadQhacqGHRaWkcaaIXaGaaG4naiabgUcaRiaaiE % dacaWGPbaabaGaamOEaiabgkHiTiaaikdacqGHsislcaWGPbaaaaGa % ay5bSlaawIa7aiabg2da9iaaigdacaaIZaGaeyO0H49aaqWaaeaada % qadaqaaiaaigdacaaIYaGaeyOeI0IaaGynaiaadMgaaiaawIcacaGL % PaaacaWG6bGaey4kaSIaaGymaiaaiEdacqGHRaWkcaaI3aGaamyAaa % Gaay5bSlaawIa7aiabg2da9iaaigdacaaIZaWaaqWaaeaacaWG6bGa % eyOeI0IaaGOmaiabgkHiTiaadMgaaiaawEa7caGLiWoaaaa!66D8! \left| {\frac{{\left( {12 - 5i} \right)z + 17 + 7i}}{{z - 2 - i}}} \right| = 13 \Rightarrow \left| {\left( {12 - 5i} \right)z + 17 + 7i} \right| = 13\left| {z - 2 - i} \right|\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aaq % WaaeaacaaIXaGaaGOmaiabgkHiTiaaiwdacaWGPbaacaGLhWUaayjc % SdWaaqWaaeaacaWG6bGaey4kaSYaaSaaaeaacaaIXaGaaG4naiabgU % caRiaaiEdacaWGPbaabaGaaGymaiaaikdacqGHsislcaaI1aGaamyA % aaaaaiaawEa7caGLiWoacqGH9aqpcaaIXaGaaG4mamaaemaabaGaam % OEaiabgkHiTiaaikdacqGHsislcaWGPbaacaGLhWUaayjcSdGaeyi1 % HS9aaqWaaeaacaWG6bGaey4kaSIaaGymaiabgUcaRiaadMgaaiaawE % a7caGLiWoacqGH9aqpdaabdaqaaiaadQhacqGHsislcaaIYaGaeyOe % I0IaamyAaaGaay5bSlaawIa7aaaa!696B! \Leftrightarrow \left| {12 - 5i} \right|\left| {z + \frac{{17 + 7i}}{{12 - 5i}}} \right| = 13\left| {z - 2 - i} \right| \Leftrightarrow \left| {z + 1 + i} \right| = \left| {z - 2 - i} \right|\)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEaiabg2 % da9iaadIhacqGHRaWkcaWG5bGaamyAaiaacYcadaWadaqaamaabmaa % baGaamiEaiaacUdacaWG5baacaGLOaGaayzkaaGaeyiyIK7aaeWaae % aacaaIYaGaai4oaiaaigdaaiaawIcacaGLPaaaaiaawUfacaGLDbaa % aaa!482E! z = x + yi,\left[ {\left( {x;y} \right) \ne \left( {2;1} \right)} \right]\) ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG4bGaey4kaSIaaGymaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOm % aaaakiabgUcaRmaabmaabaGaamyEaiabgUcaRiaaigdaaiaawIcaca % GLPaaadaahaaWcbeqaaiaaikdaaaGccqGH9aqpdaqadaqaaiaadIha % cqGHsislcaaIYaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaO % Gaey4kaSYaaeWaaeaacaWG5bGaeyOeI0IaaGymaaGaayjkaiaawMca % amaaCaaaleqabaGaaGOmaaaakiabgsDiBlaaiAdacaWG4bGaey4kaS % IaaGinaiaadMhacqGHsislcaaIZaGaeyypa0JaaGimaiaaykW7caaM % c8+aaeWaaeaacaWGKbaacaGLOaGaayzkaaaaaa!5CD7! {\left( {x + 1} \right)^2} + {\left( {y + 1} \right)^2} = {\left( {x - 2} \right)^2} + {\left( {y - 1} \right)^2} \Leftrightarrow 6x + 4y - 3 = 0\,\,\left( d \right)\)

Vậy tập hợp điểm biểu diễn số phức z là đường thẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaacQ % dacaaI2aGaamiEaiabgUcaRiaaisdacaWG5bGaeyOeI0IaaG4maiab % g2da9iaaicdacaaMc8oaaa!40EA! d:6x + 4y - 3 = 0\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WG6baacaGLhWUaayjcSdWaaSbaaSqaaiGac2gacaGGPbGaaiOBaaqa % baGccqGH9aqpcaWGpbGaamytamaaBaaaleaaciGGTbGaaiyAaiaac6 % gaaeqaaOGaeyypa0JaamizamaabmaabaGaam4taiaacUdacaWGKbaa % caGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaaIYaWaaOaaaeaacaaIXa % GaaG4maaWcbeaaaOqaaiaaikdacaaI2aaaaaaa!4DAF! {\left| z \right|_{\min }} = O{M_{\min }} = d\left( {O;d} \right) = \frac{{2\sqrt {13} }}{{26}}\)

Dựa vào đồ thị hàm số ta suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadUgadaqadaqaaiaa % dIhacqGHsislcaaIXaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYa % aaaOWaaeWaaeaacaWG4bGaeyOeI0IaaGinaaGaayjkaiaawMcaaaaa % !44AC! f\left( x \right) = k{\left( {x - 1} \right)^2}\left( {x - 4} \right)\) (với k > 0)

Mặt khác \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maapehabaWaaqWaaeaacaWGRbWaaeWaaeaacaWG4bGaeyOeI0Ia % aGymaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakmaabmaaba % GaamiEaiabgkHiTiaaisdaaiaawIcacaGLPaaaaiaawEa7caGLiWoa % caWGKbGaamiEaaWcbaGaaGimaaqaaiaaisdaa0Gaey4kIipakiabg2 % da9iaaiodacaaIYaGaeyi1HSTaam4Aaiabg2da9maalaaabaGaaG4m % aiaaikdaaeaadaWdXbqaamaaemaabaWaaeWaaeaacaWG4bGaeyOeI0 % IaaGymaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakmaabmaa % baGaamiEaiabgkHiTiaaisdaaiaawIcacaGLPaaaaiaawEa7caGLiW % oacaWGKbGaamiEaaWcbaGaaGimaaqaaiaaisdaa0Gaey4kIipaaaGc % cqGH9aqpcaaI0aaaaa!6769! S = \int\limits_0^4 {\left| {k{{\left( {x - 1} \right)}^2}\left( {x - 4} \right)} \right|dx} = 32 \Leftrightarrow k = \frac{{32}}{{\int\limits_0^4 {\left| {{{\left( {x - 1} \right)}^2}\left( {x - 4} \right)} \right|dx} }} = 4\).

Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaisdadaqadaqaaiaa % dIhacqGHsislcaaIXaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYa % aaaOWaaeWaaeaacaWG4bGaeyOeI0IaaGinaaGaayjkaiaawMcaaiab % gkDiElaadAfacqGH9aqpdaWdXbqaaiaadAgadaahaaWcbeqaaiaaik % daaaGcdaqadaqaaiaadIhaaiaawIcacaGLPaaacaWGKbGaamiEaaWc % baGaaGimaaqaaiaaisdaa0Gaey4kIipakiabg2da9maapehabaWaam % WaaeaacaaI0aWaaeWaaeaacaWG4bGaeyOeI0IaaGymaaGaayjkaiaa % wMcaamaaCaaaleqabaGaaGOmaaaakmaabmaabaGaamiEaiabgkHiTi % aaisdaaiaawIcacaGLPaaaaiaawUfacaGLDbaadaahaaWcbeqaaiaa % ikdaaaGccaWGKbGaamiEaaWcbaGaaGimaaqaaiaaisdaa0Gaey4kIi % pakiabg2da9maalaaabaGaaGymaiaaiodacaaIZaGaaGymaiaaikda % cqaHapaCaeaacaaIZaGaaGynaaaaaaa!6EB4! f\left( x \right) = 4{\left( {x - 1} \right)^2}\left( {x - 4} \right) \Rightarrow V = \int\limits_0^4 {{f^2}\left( x \right)dx} = \int\limits_0^4 {{{\left[ {4{{\left( {x - 1} \right)}^2}\left( {x - 4} \right)} \right]}^2}dx} = \frac{{13312\pi }}{{35}}\). 

Gọi I(x,y,z)  thỏa mãn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaG4mamaaFi % aabaGaamysaiaadgeaaiaawEniaiabgUcaRiaaikdadaWhcaqaaiaa % dMeacaWGcbaacaGLxdcacqGHRaWkdaWhcaqaaiaadMeacaWGdbaaca % GLxdcacqGH9aqpdaWhcaqaaiaaicdaaiaawEniaiabgkDiElaadMea % daqadaqaaiabgkHiTmaalaaabaGaaGymaaqaaiaaiAdaaaGaai4oam % aalaaabaGaaGymaaqaaiaaiodaaaGaai4oamaalaaabaGaaGymaaqa % aiaaiodaaaaacaGLOaGaayzkaaaaaa!5239! 3\overrightarrow {IA} + 2\overrightarrow {IB} + \overrightarrow {IC} = \overrightarrow 0 \Rightarrow I\left( { - \frac{1}{6};\frac{1}{3};\frac{1}{3}} \right)\)

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 % da9iaaiodacaWGnbGaamyqamaaCaaaleqabaGaaGOmaaaakiabgUca % RiaaikdacaWGnbGaamOqamaaCaaaleqabaGaaGOmaaaakiabgUcaRi % aad2eacaWGdbWaaWbaaSqabeaacaaIYaaaaOGaeyypa0JaaG4mamaa % bmaabaWaa8HaaeaacaWGnbGaamysaaGaay51GaGaey4kaSYaa8Haae % aacaWGjbGaamyqaaGaay51GaaacaGLOaGaayzkaaWaaWbaaSqabeaa % caaIYaaaaOGaey4kaSIaaGOmamaabmaabaWaa8HaaeaacaWGnbGaam % ysaaGaay51GaGaey4kaSYaa8HaaeaacaWGjbGaamOqaaGaay51Gaaa % caGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaey4kaSYaaeWaae % aadaWhcaqaaiaad2eacaWGjbaacaGLxdcacqGHRaWkdaWhcaqaaiaa % dMeacaWGdbaacaGLxdcaaiaawIcacaGLPaaadaahaaWcbeqaaiaaik % daaaaaaa!64D9! P = 3M{A^2} + 2M{B^2} + M{C^2} = 3{\left( {\overrightarrow {MI} + \overrightarrow {IA} } \right)^2} + 2{\left( {\overrightarrow {MI} + \overrightarrow {IB} } \right)^2} + {\left( {\overrightarrow {MI} + \overrightarrow {IC} } \right)^2}\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0JaaG % Onaiaad2eacaWGjbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOm % amaaFiaabaGaamytaiaadMeaaiaawEniaiaac6cadaqadaqaaiaaio % dadaWhcaqaaiaadMeacaWGbbaacaGLxdcacqGHRaWkcaaIYaWaa8Ha % aeaacaWGjbGaamOqaaGaay51GaGaey4kaSYaa8HaaeaacaWGjbGaam % 4qaaGaay51GaaacaGLOaGaayzkaaGaey4kaSIaaG4maiaadMeacaWG % bbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiaadMeacaWGcb % WaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamysaiaadoeadaahaaWc % beqaaiaaikdaaaGccqGH9aqpcaaI2aGaamytaiaadMeadaahaaWcbe % qaaiaaikdaaaGccqGHRaWkdaagaaqaaiaaiodacaWGjbGaamyqamaa % CaaaleqabaGaaGOmaaaakiabgUcaRiaaikdacaWGjbGaamOqamaaCa % aaleqabaGaaGOmaaaakiabgUcaRiaadMeacaWGdbWaaWbaaSqabeaa % caaIYaaaaaqaaiaadogacaWGVbGaamOBaiaadohacaWG0baakiaawI % J-aaaa!7103! = 6M{I^2} + 2\overrightarrow {MI} .\left( {3\overrightarrow {IA} + 2\overrightarrow {IB} + \overrightarrow {IC} } \right) + 3I{A^2} + 2I{B^2} + I{C^2} = 6M{I^2} + \underbrace {3I{A^2} + 2I{B^2} + I{C^2}}_{const}\)

Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaaBa % aaleaaciGGTbGaaiyAaiaac6gaaeqaaOGaeyi1HSTaamytaiaadMea % daWgaaWcbaGaciyBaiaacMgacaGGUbaabeaaaaa!40CA! {P_{\min }} \Leftrightarrow M{I_{\min }}\) hay M là hình chiếu của I trên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WGqbaacaGLOaGaayzkaaGaeyO0H4TaamytaiaadMeadaWgaaWcbaGa % ciyBaiaacMgacaGGUbaabeaakiabg2da9iaadsgadaWadaqaaiaadM % eacaGG7aWaaeWaaeaacaWGqbaacaGLOaGaayzkaaaacaGLBbGaayzx % aaGaeyypa0ZaaSaaaeaacaaIYaaabaGaaG4maaaaaaa!49B1! \left( P \right) \Rightarrow M{I_{\min }} = d\left[ {I;\left( P \right)} \right] = \frac{2}{3}\).

Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuamaaBa % aaleaaciGGTbGaaiyAaiaac6gaaeqaaOGaeyypa0JaaGOnaiaac6ca % daqadaqaamaalaaabaGaaGOmaaqaaiaaiodaaaaacaGLOaGaayzkaa % WaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaG4maiaac6cadaWcaaqa % aiaaiEdaaeaacaaIXaGaaGOmaaaacqGHRaWkcaaIYaGaaiOlamaala % aabaGaaGynaaqaaiaaikdaaaGaey4kaSYaaSaaaeaacaaIXaGaaG4m % aaqaaiaaisdaaaGaeyypa0ZaaSaaaeaacaaI2aGaaGymaaqaaiaaiA % daaaaaaa!4F3A! {P_{\min }} = 6.{\left( {\frac{2}{3}} \right)^2} + 3.\frac{7}{{12}} + 2.\frac{5}{2} + \frac{{13}}{4} = \frac{{61}}{6}\). 

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGbbGaamOqaiaadoeacaWGebaabeaakiabg2da9maalaaa % baGaaGymaaqaaiaaiodaaaGaamizamaabmaabaGaam4qaiaacUdada % qadaqaaiaadgeacaWGcbGaamiraaGaayjkaiaawMcaaaGaayjkaiaa % wMcaaiaac6cacaWGtbWaaSbaaSqaaiabfs5aejaadgeacaWGcbGaam % iraaqabaaaaa!49FA! {V_{ABCD}} = \frac{1}{3}d\left( {C;\left( {ABD} \right)} \right).{S_{\Delta ABD}}\) và  \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGqbGaamytaiaad6eacaWGcbaabeaakiabg2da9maalaaa % baGaaGymaaqaaiaaiodaaaGaamizamaabmaabaGaamiuaiaacUdada % qadaqaaiaadgeacaWGcbGaamiraaGaayjkaiaawMcaaaGaayjkaiaa % wMcaaiaac6cacaWGtbWaaSbaaSqaaiabfs5aejaad2eacaWGobGaam % Oqaaqabaaaaa!4A40! {V_{PMNB}} = \frac{1}{3}d\left( {P;\left( {ABD} \right)} \right).{S_{\Delta MNB}}\)

Dễ thấy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamiuaiaacUdadaqadaqaaiaadgeacaWGcbGaamiraaGaayjk % aiaawMcaaaGaayjkaiaawMcaaiabg2da9maalaaabaGaaGymaaqaai % aaikdaaaGaamizamaabmaabaGaam4qaiaacUdadaqadaqaaiaadgea % caWGcbGaamiraaGaayjkaiaawMcaaaGaayjkaiaawMcaaaaa!483D! d\left( {P;\left( {ABD} \right)} \right) = \frac{1}{2}d\left( {C;\left( {ABD} \right)} \right)\)

Mặt khác \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacqqHuoarcaWGbbGaamOqaiaadseaaeqaaOGaeyypa0ZaaSaa % aeaacaaIXaaabaGaaGOmaaaacaWGKbWaaeWaaeaacaWGebGaai4oai % aadgeacaWGcbaacaGLOaGaayzkaaGaaiOlaiaadgeacaWGcbaaaa!4510! {S_{\Delta ABD}} = \frac{1}{2}d\left( {D;AB} \right).AB\) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacqqHuoarcaWGnbGaamOtaiaadkeaaeqaaOGaeyypa0ZaaSaa % aeaacaaIXaaabaGaaGOmaaaacaWGKbWaaeWaaeaacaWGobGaai4oai % aadgeacaWGcbaacaGLOaGaayzkaaGaaiOlaiaad2eacaWGcbaaaa!453C! {S_{\Delta MNB}} = \frac{1}{2}d\left( {N;AB} \right).MB\)

Mà \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamOtaiaacUdacaWGbbGaamOqaaGaayjkaiaawMcaaiabg2da % 9maalaaabaGaaGymaaqaaiaaiodaaaGaamizamaabmaabaGaamirai % aacUdacaWGbbGaamOqaaGaayjkaiaawMcaaaaa!4399! d\left( {N;AB} \right) = \frac{1}{3}d\left( {D;AB} \right)\) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiaadk % eacqGH9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaaiaadgeacaWGcbaa % aa!3BA6! MB = \frac{1}{2}AB\)

Do đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacqqHuoarcaWGnbGaamOtaiaadkeaaeqaaOGaeyypa0ZaaSaa % aeaacaaIXaaabaGaaGOnaaaacaWGtbWaaSbaaSqaaiabfs5aejaadg % eacaWGcbGaamiraaqabaGccaaMc8UaaGPaVlaaykW7daqadaqaaiaa % ikdaaiaawIcacaGLPaaaaaa!4914! {S_{\Delta MNB}} = \frac{1}{6}{S_{\Delta ABD}}\). suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGqbGaamytaiaad6eacaWGcbaabeaakiabg2da9maalaaa % baGaaGymaaqaaiaaikdaaaGaaiOlamaalaaabaGaaGymaaqaaiaaiA % daaaGaamOvaiabg2da9maalaaabaGaamOvaaqaaiaaigdacaaIYaaa % aaaa!4352! {V_{PMNB}} = \frac{1}{2}.\frac{1}{6}V = \frac{V}{{12}}\)

HD:

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4zayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaaGOmaiaadIha % caGGUaGabmOzayaafaWaaeWaaeaacaWG4bWaaWbaaSqabeaacaaIYa % aaaaGccaGLOaGaayzkaaGaeyOeI0IaaGOmaiaadIhadaahaaWcbeqa % aiaaiwdaaaGccqGHRaWkcaaI0aGaamiEamaaCaaaleqabaGaaG4maa % aakiabgkHiTiaaikdacaWG4bGaai4oaiqadEgagaqbamaabmaabaGa % amiEaaGaayjkaiaawMcaaiabg2da9iaaicdacqGHuhY2daWabaabae % qabaGaaGOmaiaadIhacqGH9aqpcaaIWaaabaGabmOzayaafaWaaeWa % aeaacaWG4bWaaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzkaaGaey % ypa0JaamiEamaaCaaaleqabaGaaGinaaaakiabgkHiTiaaikdacaWG % 4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGymaaaacaGLBbaaaa % a!6497! g'\left( x \right) = 2x.f'\left( {{x^2}} \right) - 2{x^5} + 4{x^3} - 2x;g'\left( x \right) = 0 \Leftrightarrow \left[ \begin{array}{l} 2x = 0\\ f'\left( {{x^2}} \right) = {x^4} - 2{x^2} + 1 \end{array} \right.\)Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaadIhadaahaaWcbeqaaiaaikdaaaGccqGHLjYScaaIWaaaaa!3C62! t = {x^2} \ge 0\) nên phương trình trở thành:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG0baacaGLOaGaayzkaaGaeyypa0JaamiDamaaCaaa % leqabaGaaGOmaaaakiabgkHiTiaaikdacaWG0bGaey4kaSIaaGymaa % aa!409D! f'\left( t \right) = {t^2} - 2t + 1\)

Dựa vào hình vẽ, ta thấy (*) có hai nghiệm phân biệt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iaaigdacaGG7aGaamiDaiabg2da9iaaikdacqGHshI3daWabaab % aeqabaGaamiEaiabg2da9iabgglaXkaaigdaaeaacaWG4bGaeyypa0 % JaeyySae7aaOaaaeaacaaIYaaaleqaaaaakiaawUfaaaaa!4904! t = 1;t = 2 \Rightarrow \left[ \begin{array}{l} x = \pm 1\\ x = \pm \sqrt 2 \end{array} \right.\).

Lập bảng biến thiên ta thấy  Hàm số y = g(x) có một điểm cực tiểu

HD: 

Bất phương trình trở thành: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIZaWaaWbaaSqabeaacaWG4bGaey4kaSIaaGOmaaaakiabgkHiTiaa % iodadaahaaWcbeqaamaalaaabaGaaGymaaqaaiaaikdaaaaaaaGcca % GLOaGaayzkaaWaamWaaeaacaaIZaWaaWbaaSqabeaacaWG4baaaOGa % eyOeI0IaaG4mamaaCaaaleqabaGaciiBaiaac+gacaGGNbWaaSbaaW % qaaiaaiodaaeqaaSWaaeWaaeaacaaIYaGaamyBaaGaayjkaiaawMca % aaaaaOGaay5waiaaw2faaiabgYda8iaaicdacqGHuhY2daqadaqaai % aadIhacqGHRaWkdaWcaaqaaiaaiodaaeaacaaIYaaaaaGaayjkaiaa % wMcaamaadmaabaGaamiEaiabgkHiTiGacYgacaGGVbGaai4zamaaBa % aaleaacaaIZaaabeaakmaabmaabaGaaGOmaiaad2gaaiaawIcacaGL % PaaaaiaawUfacaGLDbaacqGH8aapcaaIWaaaaa!60D2! \left( {{3^{x + 2}} - {3^{\frac{1}{2}}}} \right)\left[ {{3^x} - {3^{{{\log }_3}\left( {2m} \right)}}} \right] < 0 \Leftrightarrow \left( {x + \frac{3}{2}} \right)\left[ {x - {{\log }_3}\left( {2m} \right)} \right] < 0\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaey % OeI0YaaSaaaeaacaaIZaaabaGaaGOmaaaacqGH8aapcaWG4bGaeyip % aWJaciiBaiaac+gacaGGNbWaaSbaaSqaaiaaiodaaeqaaOGaaGOmai % aad2gaaaa!433B! \Leftrightarrow - \frac{3}{2} < x < {\log _3}2m\) mà x nhận tối đa 9 số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % iEaiabg2da9maacmaabaGaeyOeI0IaaGymaiaacUdacaaIWaGaai4o % aiaaigdacaGG7aGaaiOlaiaac6cacaGGUaGaai4oaiaaiEdaaiaawU % hacaGL9baaaaa!4574! \Rightarrow x = \left\{ { - 1;0;1;...;7} \right\}\)

Do đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaiodaaeqaaOWaaeWaaeaacaaIYaGaamyB % aaGaayjkaiaawMcaaiabgYda8iaaiIdacqGHuhY2caWGTbGaeyipaW % ZaaSaaaeaacaaIZaWaaWbaaSqabeaacaaI4aaaaaGcbaGaaGOmaaaa % cqGHijYUcaaIZaGaaGOmaiaaiIdacaaIWaGaaiilaiaaiwdaaaa!4B9C! {\log _3}\left( {2m} \right) < 8 \Leftrightarrow m < \frac{{{3^8}}}{2} \approx 3280,5\)

HD: 

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaaG4maiaadgga % caWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiaadkgaca % WG4bGaey4kaSIaam4yaaaa!4355! f'\left( x \right) = 3a{x^2} + 2bx + c\)

Hàm số đã cho không có cực trị nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGafuiLdqKbau % aadaWgaaWcbaGaamOzamaabmaabaGaamiEaaGaayjkaiaawMcaaaqa % baGccqGH9aqpcaWGIbWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaG % 4maiaadggacaWGJbGaeyizImQaaGimaiabgsDiBlaadggacaWGJbGa % eyyzIm7aaSaaaeaacaWGIbWaaWbaaSqabeaacaaIYaaaaaGcbaGaaG % 4maaaaaaa!4C6A! {\Delta '_{f\left( x \right)}} = {b^2} - 3ac \le 0 \Leftrightarrow ac \ge \frac{{{b^2}}}{3}\).Theo bất đẳng thức Cô-si ta có:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabgw % MiZkaaikdacaWGHbGaam4yaiabgUcaRiaadkgacqGHRaWkcaaIYaGa % eyyzIm7aaSaaaeaacaaIYaGaamOyamaaCaaaleqabaGaaGOmaaaaaO % qaaiaaiodaaaGaey4kaSIaamOyaiabgUcaRiaaikdacqGH9aqpdaWc % aaqaaiaaikdaaeaacaaIZaaaamaabmaabaGaamOyaiabgUcaRmaala % aabaGaaG4maaqaaiaaisdaaaaacaGLOaGaayzkaaWaaWbaaSqabeaa % caaIYaaaaOGaey4kaSYaaSaaaeaacaaIXaGaaG4maaqaaiaaiIdaaa % GaeyyzIm7aaSaaaeaacaaIXaGaaG4maaqaaiaaiIdaaaaaaa!56AA! P \ge 2ac + b + 2 \ge \frac{{2{b^2}}}{3} + b + 2 = \frac{2}{3}{\left( {b + \frac{3}{4}} \right)^2} + \frac{{13}}{8} \ge \frac{{13}}{8}\)

Dấu bằng xảy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaadkgacqGH9aqpcqGHsisldaWcaaqaaiaaiodaaeaa % caaI0aaaaaqaaiaadggacqGH9aqpcaWGJbGaeyypa0ZaaSaaaeaada % GcaaqaaiaaiodaaSqabaaakeaacaaI0aaaaaaacaGL7baaaaa!435F! \Leftrightarrow \left\{ \begin{array}{l} b = - \frac{3}{4}\\ a = c = \frac{{\sqrt 3 }}{4} \end{array} \right.\).

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadggacaWG4bWaaWba % aSqabeaacaaIYaaaaOGaey4kaSIaamOyaiaadIhacqGHRaWkcaWGJb % GaeyO0H4TabmOzayaafaWaaeWaaeaacaWG4baacaGLOaGaayzkaaGa % eyypa0JaaGOmaiaadggacaWG4bGaey4kaSIaamOyaaaa!4D18! f\left( x \right) = a{x^2} + bx + c \Rightarrow f'\left( x \right) = 2ax + b\)

Do đó giả thiết \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aae % WaaeaacaaIYaGaamyyaiaadIhacqGHRaWkcaWGIbaacaGLOaGaayzk % aaWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGinaiaadggacaWG4b % WaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGinaiaadkgacaWG4bGa % ey4kaSIaaGinaiaadogacqGH9aqpcaaI4aGaamiEamaaCaaaleqaba % GaaGOmaaaakiabgUcaRiaaisdaaaa!4F0D! \Leftrightarrow {\left( {2ax + b} \right)^2} + 4a{x^2} + 4bx + 4c = 8{x^2} + 4\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aae % WaaeaacaaI0aGaamyyamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaa % isdacaWGHbaacaGLOaGaayzkaaGaamiEamaaCaaaleqabaGaaGOmaa % aakiabgUcaRmaabmaabaGaaGinaiaadggacaWGIbGaey4kaSIaaGin % aiaadkgaaiaawIcacaGLPaaacaWG4bGaey4kaSIaamOyamaaCaaale % qabaGaaGOmaaaakiabgUcaRiaaisdacaWGJbGaeyypa0JaaGioaiaa % dIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI0aaaaa!53FB! \Leftrightarrow \left( {4{a^2} + 4a} \right){x^2} + \left( {4ab + 4b} \right)x + {b^2} + 4c = 8{x^2} + 4\)

Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaaisdacaWGHbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGin % aiaadggacqGH9aqpcaaI4aaabaGaaGinaiaadggacaWGIbGaey4kaS % IaaGinaiaadkgacqGH9aqpcaaIWaaabaGaamOyamaaCaaaleqabaGa % aGOmaaaakiabgUcaRiaaisdacaWGJbGaeyypa0JaaGinaaaacaGL7b % aacqGHuhY2daGabaabaeqabaGaamyyaiabg2da9iaaigdaaeaacaWG % IbGaeyypa0JaaGimaaqaaiaadogacqGH9aqpcaaIXaaaaiaawUhaai % abgkDiElaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqp % caWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGymaaaa!60C8! \left\{ \begin{array}{l} 4{a^2} + 4a = 8\\ 4ab + 4b = 0\\ {b^2} + 4c = 4 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = 1\\ b = 0\\ c = 1 \end{array} \right. \Rightarrow f\left( x \right) = {x^2} + 1\). Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % WadaqaaiaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGHRaWk % caWG4baacaGLBbGaayzxaaGaamizaiaadIhaaSqaaiaaicdaaeaaca % aIXaaaniabgUIiYdGccqGH9aqpdaWcaaqaaiaaiwdaaeaacaaI2aaa % aaaa!4598! \int\limits_0^1 {\left[ {f\left( x \right) + x} \right]dx} = \frac{5}{6}\).

Xếp 9 học sinh vào 9 ghế có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaacq % qHPoWvaiaawEa7caGLiWoacqGH9aqpcaaI5aGaaiyiaaaa!3D11! \left| \Omega \right| = 9!\) cách xếp.

Gọi A là biến cố: “3 học sinh lớp 10 không ngồi 3 ghế liền nhau”.

Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa0aaaeaaca % WGbbaaaaaa!36CA! \overline A \) là biến cố: “3 học sinh lớp 10 ngồi 3 ghế liền nhau”.

Xếp 3 học sinh lớp 10 và coi là một phần tử M có 3! Cách.

Xếp phần tử M cùng 6 học sinh còn lại có: 7! Cách.

Do đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaacq % qHPoWvdaWgaaWcbaWaa0aaaeaacaWGbbaaaaqabaaakiaawEa7caGL % iWoacqGH9aqpcaaIZaGaaiyiaiaac6cacaaI3aGaaiyiaiabgkDiEl % aadcfadaqadaqaamaanaaabaGaamyqaaaaaiaawIcacaGLPaaacqGH % 9aqpdaWcaaqaaiaaiodacaGGHaGaaiOlaiaaiEdacaGGHaaabaGaaG % yoaiaacgcaaaGaeyypa0ZaaSaaaeaacaaIXaaabaGaaGymaiaaikda % aaGaeyO0H4TaamiuamaabmaabaGaamyqaaGaayjkaiaawMcaaiabg2 % da9iaaigdacqGHsislcaWGqbWaaeWaaeaadaqdaaqaaiaadgeaaaaa % caGLOaGaayzkaaGaeyypa0ZaaSaaaeaacaaIXaGaaGymaaqaaiaaig % dacaaIYaaaaaaa!5E69! \left| {{\Omega _{\overline A }}} \right| = 3!.7! \Rightarrow P\left( {\overline A } \right) = \frac{{3!.7!}}{{9!}} = \frac{1}{{12}} \Rightarrow P\left( A \right) = 1 - P\left( {\overline A } \right) = \frac{{11}}{{12}}\)

HD:

Gọi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytamaabm % aabaGaamyyaiaacUdadaWcaaqaaiaaikdacaWGHbGaeyOeI0IaaG4m % aaqaaiaadggacqGHsislcaaIYaaaaaGaayjkaiaawMcaaiabgIGiop % aabmaabaGaam4qaaGaayjkaiaawMcaaiabgkDiElqadMhagaqbamaa % bmaabaGaamyyaaGaayjkaiaawMcaaiabg2da9iabgkHiTmaalaaaba % GaaGymaaqaamaabmaabaGaamyyaiabgkHiTiaaikdaaiaawIcacaGL % PaaadaahaaWcbeqaaiaaikdaaaaaaaaa!5148! M\left( {a;\frac{{2a - 3}}{{a - 2}}} \right) \in \left( C \right) \Rightarrow y'\left( a \right) = - \frac{1}{{{{\left( {a - 2} \right)}^2}}}\) ; tâm I(2;2).

Phương trình tiếp tuyến tại M là: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9iabgkHiTmaalaaabaGaaGymaaqaamaabmaabaGaamyyaiabgkHi % TiaaikdaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaOGaai % OlamaabmaabaGaamiEaiabgkHiTiaadggaaiaawIcacaGLPaaacqGH % RaWkdaWcaaqaaiaaikdacaWGHbGaeyOeI0IaaG4maaqaaiaadggacq % GHsislcaaIYaaaaaaa!4A92! y= - \frac{1}{{{{\left( {a - 2} \right)}^2}}}.\left( {x - a} \right) + \frac{{2a - 3}}{{a - 2}}\)

Tiếp tuyến d cắt x = 2 tại \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqamaabm % aabaGaaGOmaiaacUdacaaIYaGaey4kaSYaaSaaaeaacaaIYaaabaGa % amyyaiabgkHiTiaaikdaaaaacaGLOaGaayzkaaGaeyO0H4Taamysai % aadgeacqGH9aqpdaWcaaqaaiaaikdaaeaadaabdaqaaiaadggacqGH % sislcaaIYaaacaGLhWUaayjcSdaaaaaa!4A2A! A\left( {2;2 + \frac{2}{{a - 2}}} \right) \Rightarrow IA = \frac{2}{{\left| {a - 2} \right|}}\)
Tiếp tuyến d cắt y = 2 tại .\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOqamaabm % aabaGaaGOmaiaadggacqGHsislcaaIYaGaai4oaiaaikdaaiaawIca % caGLPaaacqGHshI3caWGjbGaamOqaiabg2da9iaaikdadaabdaqaai % aadggacqGHsislcaaIYaaacaGLhWUaayjcSdaaaa!486E! B\left( {2a - 2;2} \right) \Rightarrow IB = 2\left| {a - 2} \right|\)

Do đó IA.IB = 4 mà \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4qamaaBa % aaleaacqqHuoarcaWGjbGaamyqaiaadkeaaeqaaOGaeyypa0Jaamys % aiaadgeacqGHRaWkcaWGjbGaamOqaiabgUcaRiaadgeacaWGcbGaey % ypa0JaamysaiaadgeacqGHRaWkcaWGjbGaamOqaiabgUcaRmaakaaa % baGaamysaiaadgeadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWGjb % GaamOqamaaCaaaleqabaGaaGOmaaaaaeqaaaaa!4E1C! {C_{\Delta IAB}} = IA + IB + AB = IA + IB + \sqrt {I{A^2} + I{B^2}} \)

Áp dụng bất đẳng thức AM – GM, ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadMeacaWGbbGaey4kaSIaamysaiaadkeacqGHLjYScaaIYaWa % aOaaaeaacaWGjbGaamyqaiaac6cacaWGjbGaamOqaaWcbeaakiabg2 % da9iaaisdaaeaacaWGjbGaamyqamaaCaaaleqabaGaaGOmaaaakiab % gUcaRiaadMeacaWGcbWaaWbaaSqabeaacaaIYaaaaOGaeyyzImRaaG % OmaiaadMeacaWGbbGaaiOlaiaadMeacaWGcbGaeyypa0JaaGioaaaa % caGL7baacqGHshI3caWGdbWaaSbaaSqaaiabfs5aejaadMeacaWGbb % GaamOqaaqabaGccqGHLjYScaaI0aGaey4kaSIaaGOmamaakaaabaGa % aGOmaaWcbeaaaaa!5D90! \left\{ \begin{array}{l} IA + IB \ge 2\sqrt {IA.IB} = 4\\ I{A^2} + I{B^2} \ge 2IA.IB = 8 \end{array} \right. \Rightarrow {C_{\Delta IAB}} \ge 4 + 2\sqrt 2 \)

Dấu bằng xảy ra khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiaadg % eacqGH9aqpcaWGjbGaamOqaiabg2da9iaaikdacqGHuhY2daabdaqa % aiaadggacqGHsislcaaIYaaacaGLhWUaayjcSdGaeyypa0JaaGymai % abgsDiBpaadeaaeaqabeaacaWGHbGaeyypa0JaaGymaaqaaiaadgga % cqGH9aqpcaaIZaaaaiaawUfaaaaa!4E60! IA = IB = 2 \Leftrightarrow \left| {a - 2} \right| = 1 \Leftrightarrow \left[ \begin{array}{l} a = 1\\ a = 3 \end{array} \right.\). Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaabqaeaaca % WGHbGaeyypa0JaaGinaaWcbeqab0GaeyyeIuoaaaa!3AB1! \sum {a = 4} \). 

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOEaiabg2 % da9iaadIhacqGHRaWkcaWG5bGaamyAaiaaykW7caaMc8+aaeWaaeaa % caWG4bGaaiilaiaadMhacqGHiiIZcqWIDesOaiaawIcacaGLPaaaaa % a!4601! z = x + yi\,\,\left( {x,y \in R} \right)\) nên giả thiết \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aae % WaaeaacaWG4bGaey4kaSIaaGymaaGaayjkaiaawMcaamaaCaaaleqa % baGaaGOmaaaakiabgUcaRmaabmaabaGaamyEaiabgkHiTiaaigdaai % aawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGH9aqpcaaI5aaa % aa!4532! \Leftrightarrow {\left( {x + 1} \right)^2} + {\left( {y - 1} \right)^2} = 9\).

Do đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiabg2 % da9iaaikdadaGcaaqaamaabmaabaGaamiEaiabgkHiTiaaisdaaiaa % wIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWkdaqadaqaai % aadMhacqGHRaWkcaaI1aaacaGLOaGaayzkaaWaaWbaaSqabeaacaaI % YaaaaaqabaGccqGHRaWkdaGcaaqaamaabmaabaGaamiEaiabgUcaRi % aaigdaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaGccqGHRaWk % daqadaqaaiaadMhacqGHsislcaaI3aaacaGLOaGaayzkaaWaaWbaaS % qabeaacaaIYaaaaaqabaaaaa!4FB4! A = 2\sqrt {{{\left( {x - 4} \right)}^2} + {{\left( {y + 5} \right)}^2}} + \sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {y - 7} \right)}^2}} \)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaO % aaaeaadaqadaqaaiaaikdacaWG4bGaeyOeI0IaaGioaaGaayjkaiaa % wMcaamaaCaaaleqabaGaaGOmaaaakiabgUcaRmaabmaabaGaaGOmai % aadMhacqGHRaWkcaaIXaGaaGimaaGaayjkaiaawMcaamaaCaaaleqa % baGaaGOmaaaaaeqaaOGaey4kaSYaaOaaaeaadaqadaqaaiaadIhacq % GHRaWkcaaIXaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGa % ey4kaSYaaeWaaeaacaWG5bGaeyOeI0IaaG4naaGaayjkaiaawMcaam % aaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiodadaWadaqaamaabmaa % baGaamiEaiabgUcaRiaaigdaaiaawIcacaGLPaaadaahaaWcbeqaai % aaikdaaaGccqGHRaWkdaqadaqaaiaadMhacqGHsislcaaIXaaacaGL % OaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGyoaaGaay % 5waiaaw2faaaWcbeaaaaa!60D4! = \sqrt {{{\left( {2x - 8} \right)}^2} + {{\left( {2y + 10} \right)}^2}} + \sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {y - 7} \right)}^2} + 3\left[ {{{\left( {x + 1} \right)}^2} + {{\left( {y - 1} \right)}^2} - 9} \right]} \)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyypa0ZaaO % aaaeaadaqadaqaaiaaikdacaWG4bGaeyOeI0IaaGioaaGaayjkaiaa % wMcaamaaCaaaleqabaGaaGOmaaaakiabgUcaRmaabmaabaGaaGOmai % aadMhacqGHRaWkcaaIXaGaaGimaaGaayjkaiaawMcaamaaCaaaleqa % baGaaGOmaaaaaeqaaOGaey4kaSYaaOaaaeaadaqadaqaaiaaikdaca % WG4bGaey4kaSIaaGOmaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOm % aaaakiabgUcaRmaabmaabaGaaGOmaiaadMhacqGHsislcaaI1aaaca % GLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaqabaaaaa!51DB! = \sqrt {{{\left( {2x - 8} \right)}^2} + {{\left( {2y + 10} \right)}^2}} + \sqrt {{{\left( {2x + 2} \right)}^2} + {{\left( {2y - 5} \right)}^2}} \)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyyzIm7aaO % aaaeaadaqadaqaaiaaikdacaWG4bGaeyOeI0IaaGioaiabgkHiTiaa % ikdacaWG4bGaeyOeI0IaaGOmaaGaayjkaiaawMcaamaaCaaaleqaba % GaaGOmaaaakiabgUcaRmaabmaabaGaaGOmaiaadMhacqGHRaWkcaaI % XaGaaGimaiabgkHiTiaaikdacaWG5bGaey4kaSIaaGynaaGaayjkai % aawMcaamaaCaaaleqabaGaaGOmaaaaaeqaaOGaeyypa0JaaGynamaa % kaaabaGaaGymaiaaiodaaSqabaGcdaqadaqaamaakaaabaGaamyyam % aaCaaaleqabaGaaGOmaaaakiabgUcaRiaadkgadaahaaWcbeqaaiaa % ikdaaaaabeaakiabgUcaRmaakaaabaGaam4yamaaCaaaleqabaGaaG % OmaaaakiabgUcaRiaadsgadaahaaWcbeqaaiaaikdaaaaabeaakiab % gwMiZoaakaaabaWaaeWaaeaacaWGHbGaey4kaSIaam4yaaGaayjkai % aawMcaamaaCaaaleqabaGaaGOmaaaakiabgUcaRmaabmaabaGaamOy % aiabgUcaRiaadsgaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaa % aabeaaaOGaayjkaiaawMcaaaaa!69E0! \ge \sqrt {{{\left( {2x - 8 - 2x - 2} \right)}^2} + {{\left( {2y + 10 - 2y + 5} \right)}^2}} = 5\sqrt {13} \left( {\sqrt {{a^2} + {b^2}} + \sqrt {{c^2} + {d^2}} \ge \sqrt {{{\left( {a + c} \right)}^2} + {{\left( {b + d} \right)}^2}} } \right)\)

Dấu bằng xảy ra khi và chỉ khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG4bGaai4oaiaadMhaaiaawIcacaGLPaaacqGH9aqpdaqadaqaamaa % laaabaGaeyOeI0IaaGinaiabgUcaRiaaigdacaaIYaWaaOaaaeaaca % aIZaaaleqaaaGcbaGaaGymaiaaiodaaaGaai4oamaalaaabaGaaGym % aiaaiMdacqGHsislcaaIXaGaaGioamaakaaabaGaaG4maaWcbeaaaO % qaaiaaigdacaaIZaaaaaGaayjkaiaawMcaaaaa!4A44! \left( {x;y} \right) = \left( {\frac{{ - 4 + 12\sqrt 3 }}{{13}};\frac{{19 - 18\sqrt 3 }}{{13}}} \right)\).

Suy ra _{\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqamaaBa % aaleaaciGGTbGaaiyAaiaac6gaaeqaaOGaeyypa0JaaGynamaakaaa % baGaaGymaiaaiodaaSqabaGccqGHshI3caWGHbGaeyypa0JaaGynai % aacUdacaWGIbGaeyypa0JaaGymaiaaiodadaGdKaWcbaaabeGccaGL % sgcacaaIYaGaamyyaiabgUcaRiaadkgacqGH9aqpcaaIYaGaaG4maa % aa!4DD1! {A_{\min }} = 5\sqrt {13} \Rightarrow a = 5;b = 132a + b = 23\)}

HD:
Kẻ các đường sinh AA’;BB’ của hình trụ (T).

Kẻ các đường sinh AA’;BB’ của hình trụ (T).

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOvamaaBa % aaleaacaWGpbGabm4tayaafaGaamyqaiaadkeaaeqaaOGaeyypa0Za % aSaaaeaacaaIXaaabaGaaG4maaaacaWGwbWaaSbaaSqaaiaad+eaca % WGbbGabmOqayaafaGaaiOlaiqad+eagaqbaiqadgeagaqbaiaadkea % aeqaaOGaeyypa0ZaaSaaaeaacaaIXaaabaGaaG4maaaacaWGpbGabm % 4tayaafaGaaiOlamaabmaabaWaaSaaaeaacaaIXaaabaGaaGOmaaaa % caWGpbGaamyqaiaac6cacaWGpbGabmOqayaafaGaaiOlaiGacohaca % GGPbGaaiOBaiaadgeacaWGpbGabmOqayaafaaacaGLOaGaayzkaaGa % eyypa0ZaaSaaaeaacaaIXaaabaGaaG4maaaacaWGYbWaaWbaaSqabe % aacaaIZaaaaOGaci4CaiaacMgacaGGUbGaamyqaiaad+eaceWGcbGb % auaacqGHKjYOdaWcaaqaaiaaigdaaeaacaaIZaaaaiaadkhadaahaa % Wcbeqaaiaaiodaaaaaaa!6446! {V_{OO'AB}} = \frac{1}{3}{V_{OAB'.O'A'B}} = \frac{1}{3}OO'.\left( {\frac{1}{2}OA.OB'.\sin AOB'} \right) = \frac{1}{3}{r^3}\sin AOB' \le \frac{1}{3}{r^3}\)

Dấu đẳng thức xảy ra khi và chỉ khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaecaaeaaca % WGbbGaam4taiqadkeagaqbaaGaayPadaGaeyypa0JaaGyoaiaaicda % cqGHWcaSaaa!3D91! \widehat {AOB'} = 90^\circ \) hay \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4taiaadg % eacqGHLkIxceWGpbGbauaacaWGcbaaaa!3AE5! OA \bot O'B\)

Như vậy, khối tứ diện OO'AB có thể tích lớn nhất bằng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaaG4maaaacaWGYbWaaWbaaSqabeaacaaIZaaaaaaa!395C! \frac{1}{3}{r^3}\) , đạt được khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4taiaadg % eacqGHLkIxceWGpbGbauaacaWGcbaaaa!3AE5! OA \bot O'B\). Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyqayaafa % GaamOqaiabg2da9iaadkhadaGcaaqaaiaaikdaaSqabaaaaa!3A60! A'B = r\sqrt 2 \) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadk % eacqGH9aqpdaGcaaqaaiqadgeagaqbaiaadgeadaahaaWcbeqaaiaa % ikdaaaGccqGHRaWkceWGbbGbauaacaWGcbWaaWbaaSqabeaacaaIYa % aaaaqabaGccqGH9aqpcaWGYbWaaOaaaeaacaaI2aaaleqaaaaa!4167! AB= \sqrt {A'{A^2} + A'{B^2}} = r\sqrt 6 \)

Mặt cầu (S) có tâm I(1;2;1) bán kính R=3.

Do \(\Delta\)nằm trong (P) nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8Haaeaaca % WG1bWaaSbaaSqaaiabfs5aebqabaaakiaawEniaiaac6cadaWhcaqa % aiaad6gadaWgaaWcbaWaaeWaaeaacaWGqbaacaGLOaGaayzkaaaabe % aaaOGaay51GaGaeyypa0JaaGimaiabgsDiBlaaigdacqGHsislcaWG % IbGaey4kaSIaam4yaiabg2da9iaaicdacqGHuhY2caWGIbGaeyypa0 % Jaam4yaiabgUcaRiaaigdaaaa!512D! \overrightarrow {{u_\Delta }} .\overrightarrow {{n_{\left( P \right)}}} = 0 \Leftrightarrow 1 - b + c = 0 \Leftrightarrow b = c + 1\).

Mặt khác ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamWaaeaaca % WGKbWaaeWaaeaacaWGjbGaai4oamaabmaabaGaamyqaiaadkeaaiaa % wIcacaGLPaaaaiaawIcacaGLPaaaaiaawUfacaGLDbaadaahaaWcbe % qaaiaaikdaaaGccqGHRaWkdaqadaqaamaalaaabaGaamyqaiaadkea % aeaacaaIYaaaaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaki % abg2da9iaadkfadaahaaWcbeqaaiaaikdaaaGccqGHshI3caWGKbWa % aeWaaeaacaWGjbGaai4oamaabmaabaGaamyqaiaadkeaaiaawIcaca % GLPaaaaiaawIcacaGLPaaacqGH9aqpdaGcaaqaaiaadkfadaahaaWc % beqaaiaaikdaaaGccqGHsisldaqadaqaamaalaaabaGaamyqaiaadk % eaaeaacaaIYaaaaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaa % aeqaaOGaeyypa0ZaaOaaaeaacaaI1aaaleqaaaaa!5C68! {\left[ {d\left( {I;\left( {AB} \right)} \right)} \right]^2} + {\left( {\frac{{AB}}{2}} \right)^2} = {R^2} \Rightarrow d\left( {I;\left( {AB} \right)} \right) = \sqrt {{R^2} - {{\left( {\frac{{AB}}{2}} \right)}^2}} = \sqrt 5 \)

Lại có:\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamysaiaacUdacqqHuoaraiaawIcacaGLPaaacqGH9aqpdaWc % aaqaamaaemaabaWaamWaaeaadaWhcaqaaiaadMeacaWGobGaaiOlaa % Gaay51GaWaa8HaaeaacaWG1bWaaSbaaSqaaiabfs5aebqabaaakiaa % wEniaaGaay5waiaaw2faaaGaay5bSlaawIa7aaqaamaaemaabaWaa8 % HaaeaacaWG1bWaaSbaaSqaaiabfs5aebqabaaakiaawEniaaGaay5b % SlaawIa7aaaacqGH9aqpdaWcaaqaamaaemaabaWaaeWaaeaacaaIYa % Gaam4yaiabgkHiTiaaiwdacaWGIbGaai4oaiaaiwdacaGG7aGaeyOe % I0IaaGOmaaGaayjkaiaawMcaaaGaay5bSlaawIa7aaqaamaaemaaba % WaaeWaaeaacaaIXaGaai4oaiaadkgacaGG7aGaam4yaaGaayjkaiaa % wMcaaaGaay5bSlaawIa7aaaacqGH9aqpdaWcaaqaamaakaaabaWaae % WaaeaacaaIYaGaam4yaiabgkHiTiaaiwdacaWGIbaacaGLOaGaayzk % aaWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiaaiMdaaSqaba % aakeaadaGcaaqaaiaaigdacqGHRaWkcaWGIbWaaWbaaSqabeaacaaI % YaaaaOGaey4kaSIaam4yamaaCaaaleqabaGaaGOmaaaaaeqaaaaaki % abg2da9maakaaabaGaaGynaaWcbeaaaaa!7A43! d\left( {I;\Delta } \right) = \frac{{\left| {\left[ {\overrightarrow {IN.} \overrightarrow {{u_\Delta }} } \right]} \right|}}{{\left| {\overrightarrow {{u_\Delta }} } \right|}} = \frac{{\left| {\left( {2c - 5b;5; - 2} \right)} \right|}}{{\left| {\left( {1;b;c} \right)} \right|}} = \frac{{\sqrt {{{\left( {2c - 5b} \right)}^2} + 29} }}{{\sqrt {1 + {b^2} + {c^2}} }} = \sqrt 5 \)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aaS % aaaeaadaqadaqaaiaaikdacaWGJbGaeyOeI0IaaGynaiaadogacqGH % sislcaaI1aaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaey % 4kaSIaaGOmaiaaiMdaaeaacaaIXaGaey4kaSYaaeWaaeaacaWGJbGa % ey4kaSIaaGymaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaki % abgUcaRiaadogadaahaaWcbeqaaiaaikdaaaaaaOGaeyypa0JaaGyn % aiabgsDiBpaabmaabaGaaG4maiaadogacqGHRaWkcaaI1aaacaGLOa % GaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiaaiMda % cqGH9aqpcaaI1aWaaeWaaeaacaaIYaGaamiEamaaCaaaleqabaGaaG % OmaaaakiabgUcaRiaaikdacaWGJbGaey4kaSIaaGOmaaGaayjkaiaa % wMcaaiabgsDiBlaadogadaahaaWcbeqaaiaaikdaaaGccqGHsislca % aIYaGaaGimaiaadogacqGHsislcaaI0aGaaGinaiabg2da9iaaicda % cqGHuhY2daWabaabaeqabaGaam4yaiabg2da9iaaikdacaaIYaaaba % Gaam4yaiabg2da9iabgkHiTiaaikdaaaGaay5waaaaaa!78E3! \Leftrightarrow \frac{{{{\left( {2c - 5c - 5} \right)}^2} + 29}}{{1 + {{\left( {c + 1} \right)}^2} + {c^2}}} = 5 \Leftrightarrow {\left( {3c + 5} \right)^2} + 29 = 5\left( {2{x^2} + 2c + 2} \right) \Leftrightarrow {c^2} - 20c - 44 = 0 \Leftrightarrow \left[ \begin{array}{l} c = 22\\ c = - 2 \end{array} \right.\) Do \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4yaiabg6 % da+iaaicdacqGHshI3caWGJbGaeyypa0JaaGOmaiaaikdacaGG7aGa % amOyaiabg2da9iaaikdacaaIZaGaeyO0H4TaamOyaiabgUcaRiaado % gacqGH9aqpcaaI0aGaaGynaaaa!4A16! c > 0 \Rightarrow c = 22;b = 23 \Rightarrow b + c = 45\)

Số tiền gốc ban đầu gửi vào mỗi tháng là: A = 0,2x

Số tiền cả gốc và lãi sau 3 năm (36 tháng) là: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqamaaBa % aaleaacaaIXaaabeaakiabg2da9iaadgeadaqadaqaaiaaigdacqGH % RaWkcaWGYbaacaGLOaGaayzkaaWaaSaaaeaadaqadaqaaiaaigdacq % GHRaWkcaWGYbaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIZaGaaGOn % aaaakiabgkHiTiaaigdaaeaacaWGYbaaaiabg2da9iaaicdacaGGSa % GaaGOmaiaadIhadaqadaqaaiaaigdacqGHRaWkcaWGYbaacaGLOaGa % ayzkaaWaaSaaaeaadaqadaqaaiaaigdacqGHRaWkcaWGYbaacaGLOa % GaayzkaaWaaWbaaSqabeaacaaIZaGaaGOnaaaakiabgkHiTiaaigda % aeaacaWGYbaaaaaa!56D9! {A_1} = A\left( {1 + r} \right)\frac{{{{\left( {1 + r} \right)}^{36}} - 1}}{r} = 0,2x\left( {1 + r} \right)\frac{{{{\left( {1 + r} \right)}^{36}} - 1}}{r}\)

Bắt đầu từ tháng 37, số tiền gốc gửi vào ngân hàng là: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG4bGaey4kaSIaamiEaiaac6cacaaIXaGaaGimaiaacwcaaiaawIca % caGLPaaacaGGUaGaaGOmaiaaicdacaGGLaGaeyypa0JaaGimaiaacY % cacaaIYaGaaGOmaiaadIhaaaa!44DE! \left( {x + x.10\% } \right).20\% = 0,22x\)

Số tiền cả gốc và lãi sau 4 năm (48 tháng) là: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqamaaBa % aaleaacaaIXaaabeaakmaabmaabaGaaGymaiabgUcaRiaadkhaaiaa % wIcacaGLPaaadaahaaWcbeqaaiaaigdacaaIYaaaaOGaey4kaSIaaG % imaiaacYcacaaIYaGaaGOmaiaadIhadaqadaqaaiaaigdacqGHRaWk % caWGYbaacaGLOaGaayzkaaWaaSaaaeaadaqadaqaaiaaigdacqGHRa % WkcaWGYbaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIXaGaaGOmaaaa % kiabgkHiTiaaigdaaeaacaWGYbaaaiabg2da9iaaigdacaaIWaGaaG % imaiaac6cacaaIWaGaaGimaiaaicdacaGGUaGaaGimaiaaicdacaaI % Waaaaa!57C2! {A_1}{\left( {1 + r} \right)^{12}} + 0,22x\left( {1 + r} \right)\frac{{{{\left( {1 + r} \right)}^{12}} - 1}}{r} = 100.000.000\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaaG % imaiaacYcacaaIYaGaamiEamaabmaabaGaaGymaiabgUcaRiaadkha % aiaawIcacaGLPaaadaahaaWcbeqaaiaaigdacaaIZaaaaOWaaSaaae % aadaqadaqaaiaaigdacqGHRaWkcaWGYbaacaGLOaGaayzkaaWaaWba % aSqabeaacaaIZaGaaGOnaaaakiabgkHiTiaaigdaaeaacaWGYbaaai % abgUcaRiaaicdacaGGSaGaaGOmaiaaikdacaWG4bWaaeWaaeaacaaI % XaGaey4kaSIaamOCaaGaayjkaiaawMcaamaalaaabaWaaeWaaeaaca % aIXaGaey4kaSIaamOCaaGaayjkaiaawMcaamaaCaaaleqabaGaaGym % aiaaikdaaaGccqGHsislcaaIXaaabaGaamOCaaaacqGH9aqpcaaIXa % GaaGimaiaaicdacaGGUaGaaGimaiaaicdacaaIWaGaaiOlaiaaicda % caaIWaGaaGimaaaa!640B! \Leftrightarrow 0,2x{\left( {1 + r} \right)^{13}}\frac{{{{\left( {1 + r} \right)}^{36}} - 1}}{r} + 0,22x\left( {1 + r} \right)\frac{{{{\left( {1 + r} \right)}^{12}} - 1}}{r} = 100.000.000\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % iEaiabgIKi7kaaiIdacaGGUaGaaGyoaiaaiMdacaaIXaGaaiOlaiaa % iwdacaaIWaGaaGinaaaa!419C! \Rightarrow x \approx 8.991.504\)đồng

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadwhacqGH9aqpcaWG4baabaGaamizaiaadAhacqGH9aqpceWG % MbGbauaadaqadaqaaiaadIhaaiaawIcacaGLPaaacaWGKbGaamiEaa % aacaGL7baacqGHuhY2daGabaabaeqabaGaamizaiaadwhacqGH9aqp % caWGKbGaamiEaaqaaiaadAhacqGH9aqpcaWGMbWaaeWaaeaacaWG4b % aacaGLOaGaayzkaaaaaiaawUhaaiabgkDiElaadMeacqGH9aqpcaWG % 4bGaaiOlaiaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaadaabba % qaauaabeqaceaaaeaacaaIXaaabaGaaGimaaaaaiaawEa7aiabgkHi % TmaapehabaGaamOzamaabmaabaGaamiEaaGaayjkaiaawMcaaiaads % gacaWG4bGaeyypa0JaamOzamaabmaabaGaaGymaaGaayjkaiaawMca % aiabgkHiTaWcbaGaaGimaaqaaiaaigdaa0Gaey4kIipakmaapehaba % GaamOzamaabmaabaGaamiEaaGaayjkaiaawMcaaiaadsgacaWG4baa % leaacaaIWaaabaGaaGymaaqdcqGHRiI8aaaa!7506! \left\{ \begin{array}{l} u = x\\ dv = f'\left( x \right)dx \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} du = dx\\ v = f\left( x \right) \end{array} \right. \Rightarrow I = x.f\left( x \right)\left| {\begin{array}{*{20}{c}} 1\\ 0 \end{array}} \right. - \int\limits_0^1 {f\left( x \right)dx = f\left( 1 \right) - } \int\limits_0^1 {f\left( x \right)dx} \)

Thay \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGH9aqpcaaIXaaabaGaamiEaiabg2da9iaaicdaaaGa % ay5Eaaaaaa!3C8F! \left\{ \begin{array}{l} x = 1\\ x = 0 \end{array} \right.\) vào giả thiết, ta được \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaaiwdacaWGMbWaaeWaaeaacaaIWaaacaGLOaGaayzkaaGaeyOe % I0IaaG4naiaadAgadaqadaqaaiaaigdaaiaawIcacaGLPaaacqGH9a % qpcaaIWaaabaGaeyOeI0IaaG4naiaadAgadaqadaqaaiaaicdaaiaa % wIcacaGLPaaacqGHRaWkcaaI1aGaamOzamaabmaabaGaaGymaaGaay % jkaiaawMcaaiabg2da9iabgkHiTiaaiodaaaGaay5EaaGaeyi1HS9a % aiqaaqaabeqaaiaadAgadaqadaqaaiaaicdaaiaawIcacaGLPaaacq % GH9aqpdaWcaaqaaiaaiEdaaeaacaaI4aaaaaqaaiaadAgadaqadaqa % aiaaigdaaiaawIcacaGLPaaacqGH9aqpdaWcaaqaaiaaiwdaaeaaca % aI4aaaaaaacaGL7baaaaa!5D04! \left\{ \begin{array}{l} 5f\left( 0 \right) - 7f\left( 1 \right) = 0\\ - 7f\left( 0 \right) + 5f\left( 1 \right) = - 3 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} f\left( 0 \right) = \frac{7}{8}\\ f\left( 1 \right) = \frac{5}{8} \end{array} \right.\).
Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaaca % aI1aGaamOzamaabmaabaGaamiEaaGaayjkaiaawMcaaiaadsgacaWG % 4baaleaacaaIWaaabaGaaGymaaqdcqGHRiI8aOGaeyOeI0Yaa8qCae % aacaaI3aGaamOzamaabmaabaGaaGymaiabgkHiTiaadIhaaiaawIca % caGLPaaacaWGKbGaamiEaaWcbaGaaGimaaqaaiaaigdaa0Gaey4kIi % pakiabg2da9maapehabaGaaG4mamaabmaabaGaamiEamaaCaaaleqa % baGaaGOmaaaakiabgkHiTiaaikdacaWG4baacaGLOaGaayzkaaGaam % izaiaadIhaaSqaaiaaicdaaeaacaaIXaaaniabgUIiYdGccqGHshI3 % cqGHsislcaaIYaWaa8qCaeaacaWGMbWaaeWaaeaacaWG4baacaGLOa % GaayzkaaGaamizaiaadIhaaSqaaiaaicdaaeaacaaIXaaaniabgUIi % YdGccqGH9aqpcqGHsislcaaIYaGaeyO0H49aa8qCaeaacaWGMbWaae % WaaeaacaWG4baacaGLOaGaayzkaaGaamizaiaadIhaaSqaaiaaicda % aeaacaaIXaaaniabgUIiYdGccqGH9aqpcaaIXaaaaa!7787! \int\limits_0^1 {5f\left( x \right)dx} - \int\limits_0^1 {7f\left( {1 - x} \right)dx} = \int\limits_0^1 {3\left( {{x^2} - 2x} \right)dx} \Rightarrow - 2\int\limits_0^1 {f\left( x \right)dx} = - 2 \Rightarrow \int\limits_0^1 {f\left( x \right)dx} = 1\).

Do đó . \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9iaadAgadaqadaqaaiaaigdaaiaawIcacaGLPaaacqGHsislcaaI % XaGaeyypa0ZaaSaaaeaacaaI1aaabaGaaGioaaaacqGHsislcaaIXa % Gaeyypa0JaeyOeI0YaaSaaaeaacaaIZaaabaGaaGioaaaadaGdKaWc % baaabeGccaGLsgcacaWGHbGaeyypa0JaaG4maiaacUdacaWGIbGaey % ypa0JaaGioaiabgkDiElaadsfacqGH9aqpcaaIXaGaaGinaaaa!51C3! I = f\left( 1 \right) - 1 = \frac{5}{8} - 1 = - \frac{3}{8}a = 3;b = 8 \Rightarrow T = 14\)

Gọi K là trung điểm của \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadk % eacqGHshI3caWGlbWaaeWaaeaadaWcaaqaaiaadggaaeaacaaIYaaa % aiaacUdadaWcaaqaaiaadkgaaeaacaaIYaaaaiaacUdacaaIWaaaca % GLOaGaayzkaaaaaa!41D3! AB \Rightarrow K\left( {\frac{a}{2};\frac{b}{2};0} \right)\) , gọi d là đường thẳng qua K là vuông góc với mặt phẳng (Oxy), mặt phẳng trung trực của OC cắt d tại điểm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysamaabm % aabaWaaSaaaeaacaWGHbaabaGaaGOmaaaacaGG7aWaaSaaaeaacaWG % IbaabaGaaGOmaaaacaGG7aWaaSaaaeaacaWGJbaabaGaaGOmaaaaai % aawIcacaGLPaaacqGHshI3caWGjbaaaa!420C! I\left( {\frac{a}{2};\frac{b}{2};\frac{c}{2}} \right) \Rightarrow I\) là tâm mặt cầu ngoại tiếp tứ diện OABC.

Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabgU % caRiaadkgacqGHRaWkcaWGJbGaeyypa0JaaGOmaiaaicdacaaIXaGa % aGioaiabgkDiEpaalaaabaGaamyyaaqaaiaaikdaaaGaey4kaSYaaS % aaaeaacaWGIbaabaGaaGOmaaaacqGHRaWkdaWcaaqaaiaadogaaeaa % caaIYaaaaiabg2da9iaaigdacaaIWaGaaGimaiaaiMdaaaa!4B97! a + b + c = 2018 \Rightarrow \frac{a}{2} + \frac{b}{2} + \frac{c}{2} = 1009\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaam % iEamaaBaaaleaacaWGjbaabeaakiabgUcaRiaadMhadaWgaaWcbaGa % amysaaqabaGccqGHRaWkcaWG6bWaaSbaaSqaaiaadMeaaeqaaOGaey % ypa0JaaGymaiaaicdacaaIWaGaaGyoaiabgkDiElaadMeaaaa!473C! \Leftrightarrow {x_I} + {y_I} + {z_I} = 1009 \Rightarrow I\) luôn thuộc mặt phẳng (P) có phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgU % caRiaadMhacqGHRaWkcaWG6bGaeyOeI0IaaGymaiaaicdacaaIWaGa % aGyoaiabg2da9iaaicdaaaa!4050! x+ y + z - 1009 = 0\).

Suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamytaiaacUdadaqadaqaaiaadcfaaiaawIcacaGLPaaaaiaa % wIcacaGLPaaacqGH9aqpdaWcaaqaamaaemaabaGaaGymaiabgkHiTi % aaigdacaaIWaGaaGimaiaaiMdaaiaawEa7caGLiWoaaeaadaGcaaqa % aiaaiodaaSqabaaaaOGaeyypa0JaaG4maiaaiodacaaI2aWaaOaaae % aacaaIZaaaleqaaaaa!4A20! d\left( {M;\left( P \right)} \right) = \frac{{\left| {1 - 1009} \right|}}{{\sqrt 3 }} = 336\sqrt 3 \)

Số điểm cực trị của hàm số đã cho là số điểm cực trị của hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maaemaabaGaaGymaiaaicdacaWGMbWaaeWaaeaacaWG4baacaGL % OaGaayzkaaGaeyOeI0YaaSaaaeaacaaIXaGaaGymaaqaaiaaiodaaa % GaamyBamaaCaaaleqabaGaaGOmaaaakiabgUcaRmaalaaabaGaaG4m % aiaaiEdaaeaacaaIZaaaaiaad2gaaiaawEa7caGLiWoaaaa!4933! y = \left| {10f\left( x \right) - \frac{{11}}{3}{m^2} + \frac{{37}}{3}m} \right|\).

Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaigdacaaIWaGaamOz % amaabmaabaGaamiEaaGaayjkaiaawMcaaiabgkHiTmaalaaabaGaaG % ymaiaaigdaaeaacaaIZaaaaiaad2gadaahaaWcbeqaaiaaikdaaaGc % cqGHRaWkdaWcaaqaaiaaiodacaaI3aaabaGaaG4maaaacaWGTbaaaa!4885! g\left( x \right) = 10f\left( x \right) - \frac{{11}}{3}{m^2} + \frac{{37}}{3}m\) thì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabm4zayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaaGymaiaaicda % ceWGMbGbauaadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpca % aIWaaaaa!4129! g'\left( x \right) = 10f'\left( x \right) = 0\) có 2 nghiệm phân biệt.

Lại có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4zamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaaicdacqGHuhY2caWG % MbWaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0ZaaSaaaeaaca % aIXaGaaGymaaqaaiaaiodacaaIWaaaaiaad2gadaahaaWcbeqaaiaa % ikdaaaGccqGHsisldaWcaaqaaiaaiodacaaI3aaabaGaaG4maiaaic % daaaGaamyBaiaaykW7caaMc8+aaeWaaeaacaGGQaaacaGLOaGaayzk % aaaaaa!510B! g\left( x \right) = 0 \Leftrightarrow f\left( x \right) = \frac{{11}}{{30}}{m^2} - \frac{{37}}{{30}}m\,\,\left( * \right)\), để hàm số đã cho có 3 điểm cực trị thì (*) có một nghiệm đơn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aam % qaaqaabeqaamaalaaabaGaaGymaiaaigdaaeaacaaIZaGaaGimaaaa % caWGTbWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0YaaSaaaeaacaaIZa % GaaG4naaqaaiaaiodacaaIWaaaaiaad2gacqGHLjYScaaIZaaabaWa % aSaaaeaacaaIXaGaaGymaaqaaiaaiodacaaIWaaaaiaad2gadaahaa % WcbeqaaiaaikdaaaGccqGHsisldaWcaaqaaiaaiodacaaI3aaabaGa % aG4maiaaicdaaaGaamyBaiabgsMiJkabgkHiTiaaigdaaaGaay5waa % Gaeyi1HS9aamqaaqaabeqaaiaad2gacqGHLjYScaaI1aaabaGaamyB % aiabgsMiJkabgkHiTmaalaaabaGaaGymaiaaiIdaaeaacaaIXaGaaG % ymaaaaaeaadaWcaaqaaiaaigdacaaI1aaabaGaaGymaiaaigdaaaGa % eyizImQaamyBaiabgsMiJkaaikdaaaGaay5waaaaaa!6842! \Leftrightarrow \left[ \begin{array}{l} \frac{{11}}{{30}}{m^2} - \frac{{37}}{{30}}m \ge 3\\ \frac{{11}}{{30}}{m^2} - \frac{{37}}{{30}}m \le - 1 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} m \ge 5\\ m \le - \frac{{18}}{{11}}\\ \frac{{15}}{{11}} \le m \le 2 \end{array} \right.\).

Kết hợp \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaad2gacqGHiiIZcqWIKeIOaeaacaWGTbGaeyicI48aamWaaeaa % cqGHsislcaaIYaGaaGimaiaacUdacaaIYaGaaGimaaGaay5waiaaw2 % faaaaacaGL7baacqGHshI3aaa!465F! \left\{ \begin{array}{l} m \in Z \\ m \in \left[ { - 20;20} \right] \end{array} \right. \Rightarrow \) có 36 giá trị của m

Chia hai vế của giả thiết cho \(x^2\) ta được \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaG4maiaadM % hacaGGUaWaamWaaeaacaaIXaGaey4kaSYaaOaaaeaacaaIXaGaey4k % aSYaaeWaaeaacaaIZaGaamyEaaGaayjkaiaawMcaamaaCaaaleqaba % GaaGOmaaaaaeqaaaGccaGLBbGaayzxaaGaeyypa0ZaaSaaaeaacaaI % YaaabaGaamiEaaaacqGHRaWkdaWcaaqaaiaaikdadaGcaaqaaiaadI % hadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaI0aaaleqaaaGcbaGa % amiEamaaCaaaleqabaGaaGOmaaaaaaGccqGHuhY2caaIZaGaamyEai % abgUcaRiaaiodacaWG5bGaaiOlamaakaaabaGaaGymaiabgUcaRmaa % bmaabaGaaG4maiaadMhaaiaawIcacaGLPaaadaahaaWcbeqaaiaaik % daaaaabeaakiabg2da9maalaaabaGaaGOmaaqaaiaadIhaaaGaey4k % aSYaaSaaaeaacaaIYaaabaGaamiEaaaacaGGUaWaaOaaaeaacaaIXa % Gaey4kaSYaaeWaaeaadaWcaaqaaiaaikdaaeaacaWG4baaaaGaayjk % aiaawMcaamaaCaaaleqabaGaaGOmaaaaaeqaaaaa!6553! 3y.\left[ {1 + \sqrt {1 + {{\left( {3y} \right)}^2}} } \right] = \frac{2}{x} + \frac{{2\sqrt {{x^2} + 4} }}{{{x^2}}} \Leftrightarrow 3y + 3y.\sqrt {1 + {{\left( {3y} \right)}^2}} = \frac{2}{x} + \frac{2}{x}.\sqrt {1 + {{\left( {\frac{2}{x}} \right)}^2}} \)

Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiDaaGaayjkaiaawMcaaiabg2da9iaadshacqGHRaWkcaWG % 0bWaaOaaaeaacaaIXaGaey4kaSIaamiDamaaCaaaleqabaGaaGOmaa % aaaeqaaaaa!40C9! f\left( t \right) = t + t\sqrt {1 + {t^2}} \) trên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIWaGaai4oaiabgUcaRiabg6HiLcGaayjkaiaawMcaaaaa!3B48! \left( {0; + \infty } \right)\) , có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG0baacaGLOaGaayzkaaGaeyypa0JaaGymaiabgUca % RmaakaaabaGaaGymaiabgUcaRiaadshadaahaaWcbeqaaiaaikdaaa % aabeaakiabgUcaRmaalaaabaGaamiDamaaCaaaleqabaGaaGOmaaaa % aOqaamaakaaabaGaaGymaiabgUcaRiaadshadaahaaWcbeqaaiaaik % daaaaabeaaaaGccqGH+aGpcaaIWaaaaa!47E1! f'\left( t \right) = 1 + \sqrt {1 + {t^2}} + \frac{{{t^2}}}{{\sqrt {1 + {t^2}} }} > 0\).

Suy ra f(t) là hàm đồng biến trên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIWaGaai4oaiabgUcaRiabg6HiLcGaayjkaiaawMcaaaaa!3B48! \left( {0; + \infty } \right)\) mà \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaaG4maiaadMhaaiaawIcacaGLPaaacqGH9aqpcaWGMbWaaeWa % aeaadaWcaaqaaiaaikdaaeaacaWG4baaaaGaayjkaiaawMcaaiabgk % DiElaaiodacaWG5bGaeyypa0ZaaSaaaeaacaaIYaaabaGaamiEaaaa % cqGHuhY2caaIZaGaamiEaiaadMhacqGH9aqpcaaIYaaaaa!4D22! f\left( {3y} \right) = f\left( {\frac{2}{x}} \right) \Rightarrow 3y = \frac{2}{x} \Leftrightarrow 3xy = 2\).

Do đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 % da9iaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaI0aGaamiE % aiaac6cacaaIZaGaamiEaiaadMhacqGHRaWkcaaI0aGaeyypa0Jaam % iEamaaCaaaleqabaGaaG4maaaakiabgkHiTiaaiIdacaWG4bGaey4k % aSIaaGinamaaoqcaleaaaeqakiaawkziamaaxababaGaciyBaiaacM % gacaGGUbGaamiuaaWcbaWaaeWaaeaacaaIWaGaai4oaiabgUcaRiab % g6HiLcGaayjkaiaawMcaaaqabaGccqGH9aqpdaWcaaqaaiaaiodaca % aI2aGaeyOeI0IaaG4maiaaikdadaGcaaqaaiaaiAdaaSqabaaakeaa % caaI5aaaaaaa!5A2C! P = {x^3} - 4x.3xy + 4 = {x^3} - 8x + 4\Rightarrow\mathop {\min P}\limits_{\left( {0; + \infty } \right)} = \frac{{36 - 32\sqrt 6 }}{9}\) khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaaGOmamaakaaabaGaaGOnaaWcbeaaaOqaaiaaioda % aaaaaa!3A64! x = \frac{{2\sqrt 6 }}{3}\) .

Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabg2 % da9iaaiodacaaI2aGaai4oaiaadkgacqGH9aqpcqGHsislcaaIZaGa % aGOmaiaacUdacaWGJbGaeyypa0JaaGyoamaaoqcaleaaaeqakiaawk % ziamaalaaabaGaamyyaiabgUcaRiaadkgaaeaacaWGJbaaaiabg2da % 9maalaaabaGaaGinaaqaaiaaiMdaaaaaaa!49B4! a = 36;b = - 32;c = 9\Rightarrow\frac{{a + b}}{c} = \frac{4}{9}\). 

Ta có: \( \left| {{z^2} + 1} \right| = 2\left| z \right| \Leftrightarrow 2 = \left| {z + \frac{1}{z}} \right| \Leftrightarrow 4 = {\left| {z + \frac{1}{z}} \right|^2} = \left( {z + \frac{1}{z}} \right)\left( {\overline z + \frac{1}{{\overline z }}} \right)\)

\( = {\left| z \right|^2} + \frac{{{z^2} + {{\left( {\overline z } \right)}^2}}}{{{{\left| z \right|}^2}}} + \frac{1}{{{{\left| z \right|}^2}}} = \frac{{{{\left| z \right|}^4} + {{\left( {z + \overline z } \right)}^2} - 2{{\left| z \right|}^2} + 1}}{{{{\left| z \right|}^2}}}\). Khi đó \({\left| z \right|^4} - 6{\left| z \right|^2} + 1 = - {\left( {z + \overline z } \right)^2} \le 0\)

Suy ra max |z| = \( = 1+ \sqrt{2}\); min |z| = \(-1+\sqrt{2}\) . Dấu bằng xảy ra khi và chỉ khi \(z + \overline z = 0 \Rightarrow z\) là số ảo.

Khi đó, \(z_1 = (-1+ \sqrt{2})i\) ; \(z_2 = (+ \sqrt{2})i\)

\(|z_1|^2+|z_2|^2 = 6\)

Gọi K,H lần lượt là hình chiếu vuông góc của M và N trên CD.

Khi quay MN quanh CD ta được mặt xung quanh của hình trụ có bán kính đáy r =2 và chiều cao \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiAaiabg2 % da9iaaigdacqGHshI3caWGtbWaaSbaaSqaaiaaigdaaeqaaOGaeyyp % a0JaaGOmaiabec8aWjaadkhacaWGObGaeyypa0JaaGinaiabec8aWb % aa!45AB! h = 1 \Rightarrow {S_1} = 2\pi rh = 4\pi \).

Khi quay MD và NC quanh CD ta được mặt xung quanh của hình nón có đường sinh lần lượt là MD và NC, bán kính đáy r =1.

Tổng diện tích xung quanh của 2 mặt này là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uamaaBa % aaleaacaaIYaaabeaakiabg2da9iabec8aWjaadkhacaGGUaGaamyt % aiaadseacqGHRaWkcqaHapaCcaWGYbGaaiOlaiaad6eacaWGdbGaey % ypa0JaeqiWdaNaaiOlaiaaikdacaGGUaWaaeWaaeaacaWGnbGaamir % aiabgUcaRiaad6eacaWGdbaacaGLOaGaayzkaaaaaa!4E2B! {S_2} = \pi r.MD + \pi r.NC = \pi .2.\left( {MD + NC} \right)\).

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaad2 % eacqGH9aqpcaWG4bGaeyO0H4TaamOtaiaadkeacqGH9aqpcaaIXaGa % eyOeI0IaamiEaaaa!4130! AM = x \Rightarrow NB = 1 - x\) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiraiaad2 % eacqGH9aqpdaGcaaqaaiaaisdacqGHRaWkcaWG4bWaaWbaaSqabeaa % caaIYaaaaaqabaGccaGGSaGaamOtaiaadkeacqGH9aqpdaGcaaqaai % aaisdacqGHRaWkdaqadaqaaiaaigdacqGHsislcaWG4baacaGLOaGa % ayzkaaWaaWbaaSqabeaacaaIYaaaaaqabaaaaa!464B! DM = \sqrt {4 + {x^2}} ,NB = \sqrt {4 + {{\left( {1 - x} \right)}^2}} \).

Diện tích toàn phần của vật thể là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9iaaisdacqaHapaCcqGHRaWkcaaIYaGaeqiWda3aaeWaaeaadaGc % aaqaaiaaisdacqGHRaWkcaWG4bWaaWbaaSqabeaacaaIYaaaaaqaba % GccqGHRaWkdaGcaaqaaiaaisdacqGHRaWkdaqadaqaaiaaigdacqGH % sislcaWG4baacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaqaba % aakiaawIcacaGLPaaaaaa!4A83! S = 4\pi + 2\pi \left( {\sqrt {4 + {x^2}} + \sqrt {4 + {{\left( {1 - x} \right)}^2}} } \right)\)nhỏ nhất

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaOaaaeaaca % aI0aGaey4kaSIaamiEamaaCaaaleqabaGaaGOmaaaaaeqaaOGaey4k % aSYaaOaaaeaacaaI0aGaey4kaSYaaeWaaeaacaaIXaGaeyOeI0Iaam % iEaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaeqaaaaa!413C! \sqrt {4 + {x^2}} + \sqrt {4 + {{\left( {1 - x} \right)}^2}} \) nhỏ nhất.

Mặt khác \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaOaaaeaaca % aI0aGaey4kaSIaamiEamaaCaaaleqabaGaaGOmaaaaaeqaaOGaey4k % aSYaaOaaaeaacaaI0aGaey4kaSYaaeWaaeaacaaIXaGaeyOeI0Iaam % iEaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaeqaaOGaeyyz % Im7aaOaaaeaadaqadaqaaiaaikdacqGHRaWkcaaIYaaacaGLOaGaay % zkaaWaaWbaaSqabeaacaaIYaaaaOGaey4kaSYaaeWaaeaacaWG4bGa % ey4kaSIaaGymaiabgkHiTiaadIhaaiaawIcacaGLPaaadaahaaWcbe % qaaiaaikdaaaaabeaakiabg2da9maakaaabaGaaGymaiaaiEdaaSqa % baaaaa!5271! \sqrt {4 + {x^2}} + \sqrt {4 + {{\left( {1 - x} \right)}^2}} \ge \sqrt {{{\left( {2 + 2} \right)}^2} + {{\left( {x + 1 - x} \right)}^2}} = \sqrt {17} \)

(Theo bất đẳng thức \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaOaaaeaaca % WGHbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamOyamaaCaaaleqa % baGaaGOmaaaaaeqaaOGaey4kaSYaaOaaaeaacaWGJbWaaWbaaSqabe % aacaaIYaaaaOGaey4kaSIaamizamaaCaaaleqabaGaaGOmaaaaaeqa % aOGaeyyzIm7aaOaaaeaadaqadaqaaiaadggacqGHRaWkcaWGJbaaca % GLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaey4kaSYaaeWaaeaa % caWGIbGaey4kaSIaamizaaGaayjkaiaawMcaamaaCaaaleqabaGaaG % Omaaaaaeqaaaaa!4D2B! \sqrt {{a^2} + {b^2}} + \sqrt {{c^2} + {d^2}} \ge \sqrt {{{\left( {a + c} \right)}^2} + {{\left( {b + d} \right)}^2}} \);hoặc Minkowski)

Dấu bằng xảy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aaS % aaaeaacaWGHbaabaGaam4yaaaacqGH9aqpdaWcaaqaaiaadkgaaeaa % caWGKbaaaiabgsDiBpaalaaabaGaamiEaaqaaiaaigdacqGHsislca % WG4baaaiabg2da9iaaigdacqGHuhY2caWG4bGaeyypa0ZaaSaaaeaa % caaIXaaabaGaaGOmaaaacqGHshI3ciGGTbGaaiyAaiaac6gacaWGtb % Gaeyypa0JaaGinaiabec8aWjabgUcaRiaaikdacqaHapaCcaGGUaWa % aOaaaeaacaaIXaGaaG4naaWcbeaakiabgIKi7kaaiodacaaI4aGaai % ilaiaaiwdaaaa!5E9D! \Leftrightarrow \frac{a}{c} = \frac{b}{d} \Leftrightarrow \frac{x}{{1 - x}} = 1 \Leftrightarrow x = \frac{1}{2} \Rightarrow \min S = 4\pi + 2\pi .\sqrt {17} \approx 38,5\).