Bất phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciyBaiaacg % gacaGG4bWaaiWaaeaaciGGSbGaai4BaiaacEgadaWgaaWcbaGaaG4m % aaqabaGccaWG4bGaai4oaiGacYgacaGGVbGaai4zamaaBaaaleaada % WcaaqaaiaaigdaaeaacaaIYaaaaaqabaGccaWG4baacaGL7bGaayzF % aaGaeyipaWJaaG4maaaa!47C3! \max \left\{ {{{\log }_3}x;{{\log }_{\frac{1}{2}}}x} \right\} < 3\) có tập nghiệm là.

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaacq % GHsislqaaaaaaaaaWdbiabg6HiLkaacUdacaaIYaGaaG4naaWdaiaa % wIcacaGLPaaaaaa!3C45! \left( { - \infty ;27} \right)\)

B. (8;27)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaqa % aaaaaaaaWdbmaalaaabaGaaGymaaqaaiaaiIdaaaGaai4oaiaaikda % caaI3aaapaGaayjkaiaawMcaaaaa!3B74! \left( {\frac{1}{8};27} \right)\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaqa % aaaaaaaaWdbiaaikdacaaI3aGaai4oaiabgUcaRiabg6HiLcWdaiaa % wIcacaGLPaaaaaa!3C3A! \left( {27; + \infty } \right)\)

Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án

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 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaiiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaiodaaeqaaOGaamiEaabaaaaaaaaapeGa % eyyzIm7daiGacYgacaGGVbGaai4zamaaBaaaleaadaWcaaqaaiaaig % daaeaacaaIYaaaaaqabaGccaWG4bGaeyi1HSTaamiEa8qacqGHLjYS % paGaaGymaaaa!482A! lo{g_3}x \ge {\log _{\frac{1}{2}}}x \Leftrightarrow x \ge 1\). Do đó Ta xét.

TH1. Nếu 1 > x > 0 khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciyBaiaacg % gacaGG4bWaaiWaaeaaciGGSbGaai4BaiaacEgadaWgaaWcbaGaaG4m % aaqabaGccaWG4bGaai4oaiGacYgacaGGVbGaai4zamaaBaaaleaada % WcaaqaaiaaigdaaeaacaaIYaaaaaqabaGccaWG4baacaGL7bGaayzF % aaGaeyipaWJaaG4maiabgsDiBlGacYgacaGGVbGaai4zamaaBaaale % aadaWcaaqaaiaaigdaaeaacaaIYaaaaaqabaGccaWG4bGaeyipaWJa % aG4maiabgsDiBlaadIhacqGH+aGpdaWcaaqaaiaaigdaaeaacaaI4a % aaaaaa!5758! \max \left\{ {{{\log }_3}x;{{\log }_{\frac{1}{2}}}x} \right\} < 3 \Leftrightarrow {\log _{\frac{1}{2}}}x < 3 \Leftrightarrow x > \frac{1}{8}\). Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaada % WcaaqaaiaaigdaaeaacaaI4aaaaiaacUdacaaIXaaacaGLOaGaayzk % aaaaaa!3A84! \left( {\frac{1}{8};1} \right)\)

TH2. Nếu \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaabaaa % aaaaaapeGaeyyzImRaaGymaaaa!3992! x \ge 1\) khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciyBaiaacg % gacaGG4bWaaiWaaeaaciGGSbGaai4BaiaacEgadaWgaaWcbaGaaG4m % aaqabaGccaWG4bGaai4oaiGacYgacaGGVbGaai4zamaaBaaaleaada % WcaaqaaiaaigdaaeaacaaIYaaaaaqabaGccaWG4baacaGL7bGaayzF % aaGaeyipaWJaaG4maiabgsDiBlGacYgacaGGVbGaai4zamaaBaaale % aacaaIZaaabeaakiaadIhacqGH8aapcaaIZaGaeyi1HSTaamiEaiab % gYda8iaaikdacaaI3aaaaa!567A! \max \left\{ {{{\log }_3}x;{{\log }_{\frac{1}{2}}}x} \right\} < 3 \Leftrightarrow {\log _3}x < 3 \Leftrightarrow x < 27\). Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaKGeaeaaca % aIXaGaai4oaiaaikdacaaI3aaacaGLBbGaayzkaaaaaa!3ABE! \left[ {1;27} \right)\)