Bất phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaacaWG4bWaaWba % aSqabeaacaaIYaaaaOGaeyOeI0IaamiEaiabgkHiTiaaikdaaiaawI % cacaGLPaaacqGHLjYSciGGSbGaai4BaiaacEgadaWgaaWcbaGaaGim % aiaacYcacaaI1aaabeaakmaabmaabaGaamiEaiabgkHiTiaaigdaai % aawIcacaGLPaaacqGHRaWkcaaIXaaaaa!4D82! {\log _2}\left( {{x^2} - x - 2} \right) \ge {\log _{0,5}}\left( {x - 1} \right) + 1\) có tập nghiệm là:
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Lời giải:
Báo saiĐiều kiện: x > 2
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaciGGSb % Gaai4BaiaacEgadaWgaaWcbaGaaGOmaaqabaGcdaqadaqaaiaadIha % daahaaWcbeqaaiaaikdaaaGccqGHsislcaWG4bGaeyOeI0IaaGOmaa % GaayjkaiaawMcaaiabgwMiZkGacYgacaGGVbGaai4zamaaBaaaleaa % caaIWaGaaiilaiaaiwdaaeqaaOWaaeWaaeaacaWG4bGaeyOeI0IaaG % ymaaGaayjkaiaawMcaaiabgUcaRiaaigdacqGHuhY2ciGGSbGaai4B % aiaacEgadaWgaaWcbaGaaGOmaaqabaGcdaWadaqaamaabmaabaGaam % iEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaadIhacqGHsislcaaI % YaaacaGLOaGaayzkaaWaaeWaaeaacaWG4bGaeyOeI0IaaGymaaGaay % jkaiaawMcaaaGaay5waiaaw2faaiabgwMiZkaaigdacqGHuhY2daqa % daqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaWG4bGaey % OeI0IaaGOmaaGaayjkaiaawMcaamaabmaabaGaamiEaiabgkHiTiaa % igdaaiaawIcacaGLPaaacqGHsislcaaIYaGaeyyzImRaaGimaaqaai % abgsDiBlaadIhadaahaaWcbeqaaiaaiodaaaGccqGHsislcaaIYaGa % amiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaadIhacqGHLjYSca % aIWaGaeyi1HS9aamqaaqaabeqaaiaaigdacqGHsisldaGcaaqaaiaa % ikdaaSqabaGccqGHKjYOcaWG4bGaeyizImQaaGimaaqaaiaadIhacq % GHLjYScaaIXaGaey4kaSYaaOaaaeaacaaIYaaaleqaaaaakiaawUfa % aaaaaa!91AC! \begin{array}{l} {\log _2}\left( {{x^2} - x - 2} \right) \ge {\log _{0,5}}\left( {x - 1} \right) + 1 \Leftrightarrow {\log _2}\left[ {\left( {{x^2} - x - 2} \right)\left( {x - 1} \right)} \right] \ge 1 \Leftrightarrow \left( {{x^2} - x - 2} \right)\left( {x - 1} \right) - 2 \ge 0\\ \Leftrightarrow {x^3} - 2{x^2} - x \ge 0 \Leftrightarrow \left[ \begin{array}{l} 1 - \sqrt 2 \le x \le 0\\ x \ge 1 + \sqrt 2 \end{array} \right. \end{array}\)