Tậpnghiệmcủabấtphươngtrình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaamaalaaabaGaaGymaaqaaiaaikdaaaaabeaa % kmaabmaabaGaciiBaiaac+gacaGGNbWaaSbaaSqaaiaaiAdaaeqaaO % WaaSaaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamiE % aaqaaiaadIhacqGHRaWkcaaI0aaaaaGaayjkaiaawMcaaiabgYda8i % {\log _{\frac{1}{2}}}\left( {{{\log }_6}\frac{{{x^2} + x}}{{x + 4}}} \right) < 0\) là

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maabmaabaGaeyOeI0IaaGinaiaacUdacqGHsislcaaIZaaacaGL % OaGaayzkaaGaeyOkIG8aaKGeaeaacaaI4aGaaGPaVlaacUdacqGHRa % WkcqGHEisPaiaawUfacaGLPaaaaaa!4640! S = \left( { - 4; - 3} \right) \cup \left[ {8\,; + \infty } \right)\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maajibabaGaaGioaiaaykW7caGG7aGaey4kaSIaeyOhIukacaGL % BbGaayzkaaaaaa!3F04! S = \left[ {8\,; + \infty } \right)\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maabmaabaGaeyOeI0IaeyOhIuQaaGPaVlaacUdacqGHsislcaaI % 0aaacaGLOaGaayzkaaGaeyOkIG8aaeWaaeaacqGHsislcaaIZaGaaG % PaVlaacUdacaqGGaGaaGioaaGaayjkaiaawMcaaaaa!482F! S = \left( { - \infty \,; - 4} \right) \cup \left( { - 3\,;{\rm{ }}8} \right)\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maabmaabaGaeyOeI0IaaGinaiaacUdacqGHsislcaaIZaaacaGL % OaGaayzkaaGaeyOkIG8aaeWaaeaacaaI4aGaaGPaVlaacUdacqGHRa % WkcqGHEisPaiaawIcacaGLPaaaaaa!45F6! S = \left( { - 4; - 3} \right) \cup \left( {8\,; + \infty } \right)\)

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Môn: Toán Lớp 12

Chủ đề: Hàm Số Lũy Thừa Hàm Số Mũ Và Hàm Số Lôgarit

Bài: Hàm số lũy thừa

Lời giải:

Báo sai

Ta có : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGPaVlGacY % gacaGGVbGaai4zamaaBaaaleaadaWcaaqaaiaaigdaaeaacaaIYaaa % aaqabaGcdaqadaqaaiGacYgacaGGVbGaai4zamaaBaaaleaacaaI2a % aabeaakmaalaaabaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUca % RiaadIhaaeaacaWG4bGaey4kaSIaaGinaaaaaiaawIcacaGLPaaacq % GH8aapcaaIWaGaaeiiaiabgsDiBlaabccaciGGSbGaai4BaiaacEga % daWgaaWcbaGaaGOnaaqabaGcdaqadaqaamaalaaabaGaamiEamaaCa % aaleqabaGaaGOmaaaakiabgUcaRiaadIhaaeaacaWG4bGaey4kaSIa % aGinaaaaaiaawIcacaGLPaaacqGH+aGpcaaIXaaaaa!5AC4! \,{\log _{\frac{1}{2}}}\left( {{{\log }_6}\frac{{{x^2} + x}}{{x + 4}}} \right) < 0{\rm{ }} \Leftrightarrow {\rm{ }}{\log _6}\left( {\frac{{{x^2} + x}}{{x + 4}}} \right) > 1\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aaS % aaaeaacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamiEaaqa % aiaadIhacqGHRaWkcaaI0aaaaiabg6da+iaaiAdacqGHuhY2daWcaa % qaaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGHsislcaaI1aGaamiE % aiabgkHiTiaaikdacaaI0aaabaGaamiEaiabgUcaRiaaisdaaaGaey % Opa4JaaGimaiabgsDiBlaadIhacqGHiiIZdaqadaqaaiabgkHiTiaa % isdacaGG7aGaeyOeI0IaaG4maaGaayjkaiaawMcaaiabgQIiipaabm % aabaGaaGioaiaacUdacqGHRaWkcqGHEisPaiaawIcacaGLPaaacaGG % Uaaaaa!6087! \Leftrightarrow \frac{{{x^2} + x}}{{x + 4}} > 6 \Leftrightarrow \frac{{{x^2} - 5x - 24}}{{x + 4}} > 0 \Leftrightarrow x \in \left( { - 4; - 3} \right) \cup \left( {8; + \infty } \right).\)