Cho hàm số bậc ba \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9iaadggacaWG4bWaaWba % aSqabeaacaaIZaaaaOGaey4kaSIaamOyaiaadIhadaahaaWcbeqaai % aaikdaaaGccqGHRaWkcaWGJbGaamiEaiabgUcaRiaadsgaaaa!458C! f\left( x \right) = a{x^3} + b{x^2} + cx + d\) có đồ thị như hình vẽ bên. Giá trị nhỏ nhất của biểu thức \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiuaiabg2 % da9iaadggadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaWGJbWaaWba % aSqabeaacaaIYaaaaOGaey4kaSIaamOyaiabgUcaRiaaikdaaaa!3FCB! P = {a^2} + {c^2} + b + 2\).
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Lời giải:
Báo saiHD:
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 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaadkgacqGH9aqpcqGHsisldaWcaaqaaiaaiodaaeaa % caaI0aaaaaqaaiaadggacqGH9aqpcaWGJbGaeyypa0ZaaSaaaeaada % GcaaqaaiaaiodaaSqabaaakeaacaaI0aaaaaaacaGL7baaaaa!435F! \Leftrightarrow \left\{ \begin{array}{l} b = - \frac{3}{4}\\ a = c = \frac{{\sqrt 3 }}{4} \end{array} \right.\).
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
Tuyển chọn số 5