Tập nghiệm của bất phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac6 % gadaWadaqaamaabmaabaGaamiEaiabgkHiTiaaigdaaiaawIcacaGL % PaaadaqadaqaaiaadIhacqGHsislcaaIYaaacaGLOaGaayzkaaWaae % WaaeaacaWG4bGaeyOeI0IaaG4maaGaayjkaiaawMcaaiabgUcaRiaa % igdaaiaawUfacaGLDbaacqGH+aGpcaaIWaaaaa!49B6! \ln \left[ {\left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) + 1} \right] > 0\) là

A. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIXaGaai4oaiaaikdaaiaawIcacaGLPaaacqGHQicYdaqadaqaaiaa % iodacaGG7aGaey4kaSIaeyOhIukacaGLOaGaayzkaaGaaiOlaaaa!415D! \left( {1;2} \right) \cup \left( {3; + \infty } \right).\)

B. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaacq% GHsislcqGHEisPcaGG7aGaaGymaaGaayjkaiaawMcaaiabgMIihpaa % bmaabaGaaGOmaiaacUdacaaIZaaacaGLOaGaayzkaaGaaiOlaaaa!4166! \left( { - \infty ;1} \right) \cap \left( {2;3} \right).\)

C. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIXaGaai4oaiaaikdaaiaawIcacaGLPaaacqGHPiYXdaqadaqaaiaa % iodacaGG7aGaey4kaSIaeyOhIukacaGLOaGaayzkaaGaaiOlaaaa!415B! \left( {1;2} \right) \cap \left( {3; + \infty } \right).\)

D. \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaacq % GHsislcqGHEisPcaGG7aGaaGymaaGaayjkaiaawMcaaiabgQIiipaa % bmaabaGaaGOmaiaacUdacaaIZaaacaGLOaGaayzkaaGaaiOlaaaa!4168! \left( { - \infty ;1} \right) \cup \left( {2;3} \right).\)

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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

Điều kiện: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % WG4bGaeyOeI0IaaGymaaGaayjkaiaawMcaamaabmaabaGaamiEaiab % gkHiTiaaikdaaiaawIcacaGLPaaadaqadaqaaiaadIhacqGHsislca % aIZaaacaGLOaGaayzkaaGaey4kaSIaaGymaiabg6da+iaaicdaaaa!45E0! \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) + 1 > 0\)

Khi đó bpt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aae % WaaeaacaWG4bGaeyOeI0IaaGymaaGaayjkaiaawMcaamaabmaabaGa % amiEaiabgkHiTiaaikdaaiaawIcacaGLPaaadaqadaqaaiaadIhacq % GHsislcaaIZaaacaGLOaGaayzkaaGaey4kaSIaaGymaiabg6da+iaa % igdaaaa!483D! \Leftrightarrow \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) + 1 > 1\), do đó điều kiện của bất phương trình luôn thỏa.

Ta có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aae % WaaeaacaWG4bGaeyOeI0IaaGymaaGaayjkaiaawMcaamaabmaabaGa % amiEaiabgkHiTiaaikdaaiaawIcacaGLPaaadaqadaqaaiaadIhacq % GHsislcaaIZaaacaGLOaGaayzkaaGaey4kaSIaaGymaiabg6da+iaa % igdacqGHuhY2daqadaqaaiaadIhacqGHsislcaaIXaaacaGLOaGaay % zkaaWaaeWaaeaacaWG4bGaeyOeI0IaaGOmaaGaayjkaiaawMcaamaa % bmaabaGaamiEaiabgkHiTiaaiodaaiaawIcacaGLPaaacqGH+aGpca % aIWaGaeyi1HS9aamqaaqaabeqaaiaaigdacqGH8aapcaWG4bGaeyip % aWJaaGOmaaqaaiaadIhacqGH+aGpcaaIZaaaaiaawUfaaaaa!6384! \Leftrightarrow \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) + 1 > 1 \Leftrightarrow \left( {x - 1} \right)\left( {x - 2} \right)\left( {x - 3} \right) > 0 \Leftrightarrow \left[ \begin{array}{l} 1 < x < 2\\ x > 3 \end{array} \right.\)

Vậy tập nghiệm của bất phương trình là \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9maabmaabaGaaGymaiaacUdacaaIYaaacaGLOaGaayzkaaGaeyOk % IG8aaeWaaeaacaaIZaGaai4oaiabgUcaRiabg6HiLcGaayjkaiaawM % caaaaa!4289! S = \left( {1;2} \right) \cup \left( {3; + \infty } \right)\)