Cho các số thực dương x;y thỏa mãn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaG4maiaadI % hadaahaaWcbeqaaiaaikdaaaGccaWG5bWaaeWaaeaacaaIXaGaey4k % aSYaaOaaaeaacaaI5aGaamyEamaaCaaaleqabaGaaGOmaaaakiabgU % caRiaaigdaaSqabaaakiaawIcacaGLPaaacqGH9aqpcaaIYaGaamiE % aiabgUcaRiaaikdadaGcaaqaaiaadIhadaahaaWcbeqaaiaaikdaaa % GccqGHRaWkcaaI0aaaleqaaaaa!4942! 3{x^2}y\left( {1 + \sqrt {9{y^2} + 1} } \right) = 2x + 2\sqrt {{x^2} + 4} \). Giá trị nhỏ nhất của biểu thức \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 % da9iaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIXaGaaGOm % aiaadIhadaahaaWcbeqaaiaaikdaaaGccaWG5bGaey4kaSIaaGinaa % aa!40B1! P = {x^3} - 12{x^2}y + 4\) là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGHbGaey4kaSIaamOyamaakaaabaGaaGOnaaWcbeaaaOqaaiaadoga % aaWaaeWaaeaacaWGHbGaaiilaiaadkgacaGGSaGaam4yaiabgIGiol % ablssiIcGaayjkaiaawMcaaaaa!4319! \frac{{a + b\sqrt 6 }}{c}\left( {a,b,c \in Z} \right )\) . Tính \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGHbGaey4kaSIaamOyaaqaaiaadogaaaaaaa!399A! \frac{{a + b}}{c}\).
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Lời giải:
Báo saiChia 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 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyyaiabg2 % da9iaaiodacaaI2aGaai4oaiaadkgacqGH9aqpcqGHsislcaaIZaGa % aGOmaiaacUdacaWGJbGaeyypa0JaaGyoamaaoqcaleaaaeqakiaawk % ziamaalaaabaGaamyyaiabgUcaRiaadkgaaeaacaWGJbaaaiabg2da % 9maalaaabaGaaGinaaqaaiaaiMdaaaaaaa!49B4! a = 36;b = - 32;c = 9\Rightarrow\frac{{a + b}}{c} = \frac{4}{9}\).
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
Tuyển chọn số 5