Tập nghiệm của bất phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaamaalaaabaGaaGymaaqaaiaaiodaaaaabeaa % kmaabmaabaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiA % dacaWG4bGaey4kaSIaaGynaaGaayjkaiaawMcaaiabgUcaRiGacYga % caGGVbGaai4zamaaBaaaleaacaaIZaaabeaakmaabmaabaGaamiEai % abgkHiTiaaigdaaiaawIcacaGLPaaacqGHLjYScaaIWaaaaa!4D98! {\log _{\frac{1}{3}}}\left( {{x^2} - 6x + 5} \right) + {\log _3}\left( {x - 1} \right) \ge 0\) là:
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaamaalaaabaGaaGymaaqaaiaaiodaaaaabeaa % kmaabmaabaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiA % dacaWG4bGaey4kaSIaaGynaaGaayjkaiaawMcaaiabgUcaRiGacYga % caGGVbGaai4zamaaBaaaleaacaaIZaaabeaakmaabmaabaGaamiEai % abgkHiTiaaigdaaiaawIcacaGLPaaacqGHLjYScaaIWaGaeyi1HSTa % ciiBaiaac+gacaGGNbWaaSbaaSqaaiaaiodaaeqaaOWaaeWaaeaaca % WG4bGaeyOeI0IaaGymaaGaayjkaiaawMcaaiabgwMiZkGacYgacaGG % VbGaai4zamaaBaaaleaacaaIZaaabeaakmaabmaabaGaamiEamaaCa % aaleqabaGaaGOmaaaakiabgkHiTiaaiAdacaWG4bGaey4kaSIaaGyn % aaGaayjkaiaawMcaaiabgsDiBpaaceaaeaqabeaacaWG4bWaaWbaaS % qabeaacaaIYaaaaOGaeyOeI0IaaGOnaiaadIhacqGHRaWkcaaI1aGa % eyOpa4JaaGimaaqaaiaadIhacqGHsislcaaIXaGaeyyzImRaamiEam % aaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiAdacaWG4bGaey4kaSIa % aGynaaaacaGL7baaaaa!7B52! {\log _{\frac{1}{3}}}\left( {{x^2} - 6x + 5} \right) + {\log _3}\left( {x - 1} \right) \ge 0 \Leftrightarrow {\log _3}\left( {x - 1} \right) \ge {\log _3}\left( {{x^2} - 6x + 5} \right) \Leftrightarrow \left\{ \begin{array}{l} {x^2} - 6x + 5 > 0\\ x - 1 \ge {x^2} - 6x + 5 \end{array} \right.\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaadIhacqGH8aapcaaIXaGaeyikIOTaamiEaiabg6da % +iaaiwdaaeaacaaIXaGaeyizImQaamiEaiabgsMiJkaaiAdaaaGaay % 5EaaGaeyi1HSTaaGynaiabgYda8iaadIhacqGHKjYOcaaI2aaaaa!4E13! \Leftrightarrow \left\{ \begin{array}{l} x < 1 \vee x > 5\\ 1 \le x \le 6 \end{array} \right. \Leftrightarrow 5 < x \le 6\)