Tập nghiệm của bất phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaai4BaiaacE % gadaWgaaWcbaGaaGOmaaqabaGccaGGOaGaamiEamaaCaaaleqabaGa % aGOmaaaakiabgkHiTiaaiodacaWG4bGaey4kaSIaaGymaiaacMcacq % GHKjYOcaaIWaaaaa!42C0! lo{g_2}({x^2} - 3x + 1) \le 0\) là ;

BPT \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI % ZaGaamiEaiabgUcaRiaaigdacqGH+aGpcaaIWaaabaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOGaaiikaiaadIhadaahaaWc % beqaaiaaikdaaaGccqGHsislcaaIZaGaamiEaiabgUcaRiaaigdaca % GGPaGaeyizImQaaGimaaaacaGL7baacqGHuhY2daGabaabaeqabaGa % amiEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiodacaWG4bGaey % 4kaSIaaGymaiabg6da+iaaicdaaeaacaWG4bWaaWbaaSqabeaacaaI % YaaaaOGaeyOeI0IaaG4maiaadIhacqGHRaWkcaaIXaGaeyizImQaaG % ymaaaacaGL7baacqGHuhY2daGabaabaeqabaGaamiEamaaCaaaleqa % baGaaGOmaaaakiabgkHiTiaaiodacaWG4bGaey4kaSIaaGymaiabg6 % da+iaaicdaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0Ia % aG4maiaadIhacqGHRaWkcaaIXaGaeyizImQaaGymaaaacaGL7baaaa % a!7753! \Leftrightarrow \left\{ \begin{array}{l} {x^2} - 3x + 1 > 0\\ {\log _2}({x^2} - 3x + 1) \le 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {x^2} - 3x + 1 > 0\\ {x^2} - 3x + 1 \le 1 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {x^2} - 3x + 1 > 0\\ {x^2} - 3x + 1 \le 1 \end{array} \right.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aai % qaaqaabeqaaiaadIhacqGH8aapdaWcaaqaaiaaiodacqGHsisldaGc % aaqaaiaaiwdaaSqabaaakeaacaaIYaaaaiabgIIiAlaadIhacqGH+a % GpdaWcaaqaaiaaiodacqGHRaWkdaGcaaqaaiaaiwdaaSqabaaakeaa % caaIYaaaaaqaaiaaicdacqGHKjYOcaWG4bGaeyizImQaaG4maaaaca % GL7baacqGHuhY2caWG4bGaeyicI48aaKGeaeaacaaIWaGaai4oamaa % laaabaGaaG4maiabgkHiTmaakaaabaGaaGynaaWcbeaaaOqaaiaaik % daaaaacaGLBbGaayzkaaGaeyOkIG8aaKamaeaadaWcaaqaaiaaioda % cqGHRaWkdaGcaaqaaiaaiwdaaSqabaaakeaacaaIYaaaaiaacUdaca % aIZaaacaGLOaGaayzxaaaaaa!5F8E! \Leftrightarrow \left\{ \begin{array}{l} x < \frac{{3 - \sqrt 5 }}{2} \vee x > \frac{{3 + \sqrt 5 }}{2}\\ 0 \le x \le 3 \end{array} \right. \Leftrightarrow x \in \left[ {0;\frac{{3 - \sqrt 5 }}{2}} \right) \cup \left( {\frac{{3 + \sqrt 5 }}{2};3} \right]\)