Tập nghiệm của bất phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaisdaaeqaaOWaaeWaaeaacaaIYaGaamiE % amaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiodacaWG4bGaey4kaS % IaaGymaaGaayjkaiaawMcaaiabg6da+iGacYgacaGGVbGaai4zamaa % BaaaleaacaaIYaaabeaakmaabmaabaGaaGOmaiaadIhacqGHRaWkca % aIXaaacaGLOaGaayzkaaaaaa!4BCF! {\log _4}\left( {2{x^2} + 3x + 1} \right) > {\log _2}\left( {2x + 1} \right)\) là:
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Lời giải:
Báo saiĐiều kiện: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaaikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaG4m % aiaadIhacqGHRaWkcaaIXaGaeyOpa4JaaGimaaqaaiaaikdacaWG4b % Gaey4kaSIaaGymaiabg6da+iaaicdaaaGaay5EaaGaeyi1HS9aaiqa % aqaabeqaaiaadIhacqGH8aapcqGHsislcaaIXaGaeyikIOTaamiEai % abg6da+iabgkHiTmaalaaabaGaaGymaaqaaiaaikdaaaaabaGaamiE % aiabg6da+iabgkHiTmaalaaabaGaaGymaaqaaiaaikdaaaaaaiaawU % haaiabgsDiBlaadIhacqGH+aGpcqGHsisldaWcaaqaaiaaigdaaeaa % caaIYaaaaiaac6caaaa!5E23! \left\{ \begin{array}{l} 2{x^2} + 3x + 1 > 0\\ 2x + 1 > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x < - 1 \vee x > - \frac{1}{2}\\ x > - \frac{1}{2} \end{array} \right. \Leftrightarrow x > - \frac{1}{2}.\)
Ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaisdaaeqaaOWaaeWaaeaacaaIYaGaamiE % amaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaiodacaWG4bGaey4kaS % IaaGymaaGaayjkaiaawMcaaiabg6da+iGacYgacaGGVbGaai4zamaa % BaaaleaacaaIYaaabeaakmaabmaabaGaaGOmaiaadIhacqGHRaWkca % aIXaaacaGLOaGaayzkaaGaeyi1HSTaciiBaiaac+gacaGGNbWaaSba % aSqaaiaaisdaaeqaaOWaaeWaaeaacaaIYaGaamiEamaaCaaaleqaba % GaaGOmaaaakiabgUcaRiaaiodacaWG4bGaey4kaSIaaGymaaGaayjk % aiaawMcaaiabg6da+iGacYgacaGGVbGaai4zamaaBaaaleaacaaI0a % aabeaakmaabmaabaGaaGOmaiaadIhacqGHRaWkcaaIXaaacaGLOaGa % ayzkaaWaaWbaaSqabeaacaaIYaaaaaaa!64F1! {\log _4}\left( {2{x^2} + 3x + 1} \right) > {\log _2}\left( {2x + 1} \right) \Leftrightarrow {\log _4}\left( {2{x^2} + 3x + 1} \right) > {\log _4}{\left( {2x + 1} \right)^2}\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaaG % OmaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIZaGaamiE % aiabgUcaRiaaigdacqGH+aGpcaaI0aGaamiEamaaCaaaleqabaGaaG % OmaaaakiabgUcaRiaaisdacaWG4bGaey4kaSIaaGymaiabgsDiBlaa % ikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamiEaiabgY % da8iaaicdacqGHuhY2cqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaa % aiabgYda8iaadIhacqGH8aapcaaIWaGaaiOlaaaa!590B! \Leftrightarrow 2{x^2} + 3x + 1 > 4{x^2} + 4x + 1 \Leftrightarrow 2{x^2} + x < 0 \Leftrightarrow - \frac{1}{2} < x < 0.\) (thỏa mãn điều kiện)
Vậy tập nghiệm của bất phương trình đã cho là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maabmaabaGaeyOeI0YaaSaaaeaacaaIXaaabaGaaGOmaaaacaGG % 7aGaaGimaaGaayjkaiaawMcaaaaa!3D48! S = \left( { - \frac{1}{2};0} \right)\)