Bất phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaisdaaeqaaOWaaeWaaeaacaWG4bWaaWba % aSqabeaacaaIYaaaaOGaeyOeI0IaaG4maiaadIhaaiaawIcacaGLPa % aacqGH+aGpciGGSbGaai4BaiaacEgadaWgaaWcbaGaaGOmaaqabaGc % daqadaqaaiaaiMdacqGHsislcaWG4baacaGLOaGaayzkaaaaaa!48D8! {\log _4}\left( {{x^2} - 3x} \right) > {\log _2}\left( {9 - x} \right)\) có bao nhiêu nghiệm nguyên?
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Lời giải:
Báo saiĐiều kiện : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaIZaGaamiE % aiabg6da+iaaicdaaeaacaaI5aGaeyOeI0IaamiEaiabg6da+iaaic % daaaGaay5EaaGaeyi1HS9aaiqaaqaabeqaaiaadIhadaqadaqaaiaa % dIhacqGHsislcaaIZaaacaGLOaGaayzkaaGaeyOpa4JaaGimaaqaai % aadIhacqGH8aapcaaI5aaaaiaawUhaaiabgsDiBpaaceaaeaqabeaa % daWabaabaeqabaGaamiEaiabg6da+iaaiodaaeaacaWG4bGaeyipaW % JaaGimaaaacaGLBbaaaeaacaWG4bGaeyipaWJaaGyoaaaacaGL7baa % cqGHuhY2daWabaabaeqabaGaamiEaiabgYda8iaaicdaaeaacaaIZa % GaeyipaWJaamiEaiabgYda8iaaiMdaaaGaay5waaaaaa!666B! \left\{ \begin{array}{l} {x^2} - 3x > 0\\ 9 - x > 0 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x\left( {x - 3} \right) > 0\\ x < 9 \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} \left[ \begin{array}{l} x > 3\\ x < 0 \end{array} \right.\\ x < 9 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x < 0\\ 3 < x < 9 \end{array} \right.\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaaciGGSb % Gaai4BaiaacEgadaWgaaWcbaGaaGinaaqabaGcdaqadaqaaiaadIha % daahaaWcbeqaaiaaikdaaaGccqGHsislcaaIZaGaamiEaaGaayjkai % aawMcaaiabg6da+iGacYgacaGGVbGaai4zamaaBaaaleaacaaIYaaa % beaakmaabmaabaGaaGyoaiabgkHiTiaadIhaaiaawIcacaGLPaaacq % GHuhY2daWcaaqaaiaaigdaaeaacaaIYaaaaiGacYgacaGGVbGaai4z % amaaBaaaleaacaaIYaaabeaakmaabmaabaGaamiEamaaCaaaleqaba % GaaGOmaaaakiabgkHiTiaaiodacaWG4baacaGLOaGaayzkaaGaeyOp % a4JaciiBaiaac+gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaae % aacaaI5aGaeyOeI0IaamiEaaGaayjkaiaawMcaaaqaaiabgsDiBlGa % cYgacaGGVbGaai4zamaaBaaaleaacaaIYaaabeaakmaabmaabaGaam % iEamaaCaaaleqabaGaaGOmaaaakiabgkHiTiaaiodacaWG4baacaGL % OaGaayzkaaGaeyOpa4JaaGOmaiGacYgacaGGVbGaai4zamaaBaaale % aacaaIYaaabeaakmaabmaabaGaaGyoaiabgkHiTiaadIhaaiaawIca % caGLPaaacqGHuhY2ciGGSbGaai4BaiaacEgadaWgaaWcbaGaaGOmaa % qabaGcdaqadaqaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsisl % caaIZaGaamiEaaGaayjkaiaawMcaaiabg6da+iGacYgacaGGVbGaai % 4zamaaBaaaleaacaaIYaaabeaakmaabmaabaGaaGyoaiabgkHiTiaa % dIhaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaakeaacqGHuh % Y2caWG4bWaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaG4maiaadIha % cqGH+aGpcaaI4aGaaGymaiabgkHiTiaaigdacaaI4aGaamiEaiabgU % caRiaadIhadaahaaWcbeqaaiaaikdaaaaakeaacqGHuhY2caaIXaGa % aGynaiaadIhacqGH+aGpcaaI4aGaaGymaiabgsDiBlaadIhacqGH+a % GpdaWcaaqaaiaaiIdacaaIXaaabaGaaGymaiaaiwdaaaGaeyi1HSTa % amiEaiabg6da+maalaaabaGaaGOmaiaaiEdaaeaacaaI1aaaaaaaaa!B0EF! \begin{array}{l} {\log _4}\left( {{x^2} - 3x} \right) > {\log _2}\left( {9 - x} \right) \Leftrightarrow \frac{1}{2}{\log _2}\left( {{x^2} - 3x} \right) > {\log _2}\left( {9 - x} \right)\\ \Leftrightarrow {\log _2}\left( {{x^2} - 3x} \right) > 2{\log _2}\left( {9 - x} \right) \Leftrightarrow {\log _2}\left( {{x^2} - 3x} \right) > {\log _2}{\left( {9 - x} \right)^2}\\ \Leftrightarrow {x^2} - 3x > 81 - 18x + {x^2}\\ \Leftrightarrow 15x > 81 \Leftrightarrow x > \frac{{81}}{{15}} \Leftrightarrow x > \frac{{27}}{5} \end{array}\)
Kết hợp với điều kiện xác định ta có bất phương trình có tập nghiệm là: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIYaGaaG4naaqaaiaaiwdaaaGaeyipaWJaamiEaiabgYda8iaaiMda % aaa!3C08! \frac{{27}}{5} < x < 9\)
Mà \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgI % GiolablssiIkabgkDiElaadIhacqGHiiIZdaGadaqaaiaaiAdacaGG % 7aGaaG4naiaacUdacaaI4aaacaGL7bGaayzFaaaaaa!44BD! x \in Z \Rightarrow x \in \left\{ {6;7;8} \right\}\)
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
Tuyển chọn số 4