Tìm tất cả các giá trị thực của tham số m sao cho khoảng (2;3) thuộc tập nghiệm của bất phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaiwdaaeqaaOWaaeWaaeaacaWG4bWaaWba % aSqabeaacaaIYaaaaOGaey4kaSIaaGymaaGaayjkaiaawMcaaiabg6 % da+iGacYgacaGGVbGaai4zamaaBaaaleaacaaI1aaabeaakmaabmaa % baGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaisdacaWG4b % Gaey4kaSIaamyBaaGaayjkaiaawMcaaiabgkHiTiaaigdacaqGGaGa % aeiiaiaabccacaqGOaGaaeymaiaabMcaaaa!5122! {\log _5}\left( {{x^2} + 1} \right) > {\log _5}\left( {{x^2} + 4x + m} \right) - 1{\rm{ (1)}}\)
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiikaiaaig % dacaGGPaGaeyi1HS9aaiqaaqaabeqaaiaadIhadaahaaWcbeqaaiaa % ikdaaaGccqGHRaWkcaaIXaGaeyOpa4ZaaSaaaeaacaWG4bWaaWbaaS % qabeaacaaIYaaaaOGaey4kaSIaaGinaiaadIhacqGHRaWkcaWGTbaa % baGaaGynaaaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaS % IaaGinaiaadIhacqGHRaWkcaWGTbGaeyOpa4JaaGimaaaacaGL7baa % cqGHuhY2daGabaabaeqabaGaamyBaiabg6da+iabgkHiTiaadIhada % ahaaWcbeqaaiaaikdaaaGccqGHsislcaaI0aGaamiEaiabg2da9iaa % dAgacaGGOaGaamiEaiaacMcaaeaacaWGTbGaeyipaWJaaGinaiaadI % hadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI0aGaamiEaiabgUca % RiaaiwdacqGH9aqpcaWGNbGaaiikaiaadIhacaGGPaaaaiaawUhaaa % aa!6BE5! (1) \Leftrightarrow \left\{ \begin{array}{l} {x^2} + 1 > \frac{{{x^2} + 4x + m}}{5}\\ {x^2} + 4x + m > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} m > - {x^2} - 4x = f(x)\\ m < 4{x^2} - 4x + 5 = g(x) \end{array} \right.\)
Hệ trên thỏa mãn
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiaIiIaam % iEaiabgIGiopaabmaabaGaaGOmaiaacUdacaaIZaaacaGLOaGaayzk % aaaaaa!3D06! \forall x \in \left( {2;3} \right)\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaad2gacqGHLjYSdaWfqaqaaiaad2eacaWGHbGaamiE % aaWcbaGaaGOmaiabgYda8iaadIhacqGH8aapcaaIZaaabeaakiaadA % gacaGGOaGaamiEaiaacMcacqGH9aqpcqGHsislcaaIXaGaaGOmaiaa % bccacaqGGaGaaeiiaiaabUgacaqGObGaaeyAaiaabccacaqGGaGaam % iEaiabg2da9iaaikdaaeaacaWGTbGaeyizIm6aaCbeaeaacaWGnbGa % amyAaiaad6gaaSqaaiaaikdacqGH8aapcaWG4bGaeyipaWJaaG4maa % qabaGccaWGMbGaaiikaiaadIhacaGGPaGaeyypa0JaaGymaiaaioda % caqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGRbGaaeiAai % aabMgacaqGGaGaaeiiaiaadIhacqGH9aqpcaaIYaaaaiaawUhaaiaa % bccacqGHuhY2cqGHsislcaaIXaGaaGOmaiabgsMiJkaad2gacqGHKj % YOcaaIXaGaaG4maaaa!78AA! \Leftrightarrow \left\{ \begin{array}{l} m \ge \mathop {Max}\limits_{2 < x < 3} f(x) = - 12{\rm{ khi }}x = 2\\ m \le \mathop {Min}\limits_{2 < x < 3} f(x) = 13{\rm{ khi }}x = 2 \end{array} \right.{\rm{ }} \Leftrightarrow - 12 \le m \le 13\)