Biết \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaada % WcaaqaaiGacogacaGGVbGaai4CamaaCaaaleqabaGaaGOmaaaakiaa % dIhacqGHRaWkciGGZbGaaiyAaiaac6gacaaMc8UaamiEaiGacogaca % GGVbGaai4CaiaadIhacqGHRaWkcaaIXaaabaGaci4yaiaac+gacaGG % ZbWaaWbaaSqabeaacaaI0aaaaOGaamiEaiabgUcaRiGacohacaGGPb % GaaiOBaiaaykW7caWG4bGaci4yaiaac+gacaGGZbWaaWbaaSqabeaa % caaIZaaaaOGaamiEaaaacaWGKbGaamiEaaWcbaWaaSaaaeaacqaHap % aCaeaacaaI0aaaaaqaamaalaaabaGaeqiWdahabaGaaG4maaaaa0Ga % ey4kIipakiabg2da9iaadggacqGHRaWkcaWGIbGaciiBaiaac6gaca % aIYaGaey4kaSIaam4yaiGacYgacaGGUbWaaeWaaeaacaaIXaGaey4k % aSYaaOaaaeaacaaIZaaaleqaaaGccaGLOaGaayzkaaaaaa!6DBA! \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\cos }^2}x + \sin \,x\cos x + 1}}{{{{\cos }^4}x + \sin \,x{{\cos }^3}x}}dx} = a + b\ln 2 + c\ln \left( {1 + \sqrt 3 } \right)\), với a, b, c là các số hữu tỉ. Giá trị của abc bằng:

A. 0

B. -2

C. -4

D. -6

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

Chủ đề: Đề thi THPT QG

Môn: Toán

Lời giải:

Báo sai

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9maapehabaWaaSaaaeaaciGGJbGaai4BaiaacohadaahaaWcbeqa % aiaaikdaaaGccaWG4bGaey4kaSIaci4CaiaacMgacaGGUbGaaGPaVl % aadIhaciGGJbGaai4BaiaacohacaWG4bGaey4kaSIaaGymaaqaaiGa % cogacaGGVbGaai4CamaaCaaaleqabaGaaGinaaaakiaadIhacqGHRa % WkciGGZbGaaiyAaiaac6gacaaMc8UaamiEaiGacogacaGGVbGaai4C % amaaCaaaleqabaGaaG4maaaakiaadIhaaaGaamizaiaadIhaaSqaam % aalaaabaGaeqiWdahabaGaaGinaaaaaeaadaWcaaqaaiabec8aWbqa % aiaaiodaaaaaniabgUIiYdGccqGH9aqpdaWdXbqaamaalaaabaGaaG % ymaiabgUcaRiGacshacaGGHbGaaiOBaiaaykW7caWG4bGaey4kaSIa % aGymaiabgUcaRiGacshacaGGHbGaaiOBamaaCaaaleqabaGaaGOmaa % aakiaadIhaaeaaciGGJbGaai4BaiaacohadaahaaWcbeqaaiaaikda % aaGccaWG4bWaaeWaaeaacaaIXaGaey4kaSIaciiDaiaacggacaGGUb % GaaGPaVlaadIhaaiaawIcacaGLPaaaaaGaamizaiaadIhacqGH9aqp % daWdXbqaamaalaaabaGaciiDaiaacggacaGGUbWaaWbaaSqabeaaca % aIYaaaaOGaamiEaiabgUcaRiGacshacaGGHbGaaiOBaiaaykW7caWG % 4bGaey4kaSIaaGOmaaqaaiGacogacaGGVbGaai4CamaaCaaaleqaba % GaaGOmaaaakiaadIhadaqadaqaaiaaigdacqGHRaWkciGG0bGaaiyy % aiaac6gacaaMc8UaamiEaaGaayjkaiaawMcaaaaacaWGKbGaamiEaa % WcbaWaaSaaaeaacqaHapaCaeaacaaI0aaaaaqaamaalaaabaGaeqiW % dahabaGaaG4maaaaa0Gaey4kIipaaSqaamaalaaabaGaeqiWdahaba % GaaGinaaaaaeaadaWcaaqaaiabec8aWbqaaiaaiodaaaaaniabgUIi % Ydaaaa!ABCB! I = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\cos }^2}x + \sin \,x\cos x + 1}}{{{{\cos }^4}x + \sin \,x{{\cos }^3}x}}dx} = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{1 + \tan \,x + 1 + {{\tan }^2}x}}{{{{\cos }^2}x\left( {1 + \tan \,x} \right)}}dx = \int\limits_{\frac{\pi }{4}}^{\frac{\pi }{3}} {\frac{{{{\tan }^2}x + \tan \,x + 2}}{{{{\cos }^2}x\left( {1 + \tan \,x} \right)}}dx} } \)

Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9iGacshacaGGHbGaaiOBaiaaykW7caWG4baaaa!3D4C! t = \tan \,x\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % izaiaadshacqGH9aqpdaWcaaqaaiaaigdaaeaaciGGJbGaai4Baiaa % cohadaahaaWcbeqaaiaaikdaaaGccaWG4baaaiaadsgacaWG4baaaa!42AD! \Rightarrow dt = \frac{1}{{{{\cos }^2}x}}dx\) Đổi cận \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadIhacqGH9aqpdaWcaaqaaiabec8aWbqaaiaaisdaaaGaeyO0 % H4TaamiDaiabg2da9iaaigdaaeaacaWG4bGaeyypa0ZaaSaaaeaacq % aHapaCaeaacaaIZaaaaiabgkDiElaadshacqGH9aqpdaGcaaqaaiaa % iodaaSqabaaaaOGaay5Eaaaaaa!4A85! \left\{ \begin{array}{l} x = \frac{\pi }{4} \Rightarrow t = 1\\ x = \frac{\pi }{3} \Rightarrow t = \sqrt 3 \end{array} \right.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGH9a % qpdaWcaaqaaiaadshadaahaaWcbeqaaiaaikdaaaaakeaacaaIYaaa % aiabgUcaRiaaikdaciGGSbGaaiOBamaaemaabaGaamiDaiabgUcaRi % aaigdaaiaawEa7caGLiWoadaabbaabaeqabaWaaWbaaSqabeaadaGc % aaqaaiaaiodaaWqabaaaaaGcbaWaaSbaaSqaaiaaigdaaeqaaaaaki % aawEa7aiabg2da9maalaaabaGaaG4maaqaaiaaikdaaaGaey4kaSIa % aGOmaiGacYgacaGGUbWaaeWaaeaadaGcaaqaaiaaiodaaSqabaGccq % GHRaWkcaaIXaaacaGLOaGaayzkaaGaeyOeI0YaaSaaaeaacaaIXaaa % baGaaGOmaaaacqGHsislcaaIYaGaciiBaiaac6gacaaIYaGaeyypa0 % JaaGymaiabgkHiTiaaikdaciGGSbGaaiOBaiaaikdacqGHRaWkcaaI % YaGaciiBaiaac6gadaqadaqaaiaaigdacqGHRaWkdaGcaaqaaiaaio % daaSqabaaakiaawIcacaGLPaaaaeaacqGHshI3daGabaabaeqabaGa % amyyaiabg2da9iaaigdaaeaacaWGIbGaeyypa0JaeyOeI0IaaGOmaa % qaaiaadogacqGH9aqpcaaIYaaaaiaawUhaaiabgkDiElaadggacaWG % IbGaam4yaiabg2da9iaaigdacaGGUaGaaiikaiabgkHiTiaaikdaca % GGPaGaaiOlaiaaikdacqGH9aqpcqGHsislcaaI0aaaaaa!8005! \begin{array}{l} = \frac{{{t^2}}}{2} + 2\ln \left| {t + 1} \right|\left| \begin{array}{l} ^{\sqrt 3 }\\ _1 \end{array} \right. = \frac{3}{2} + 2\ln \left( {\sqrt 3 + 1} \right) - \frac{1}{2} - 2\ln 2 = 1 - 2\ln 2 + 2\ln \left( {1 + \sqrt 3 } \right)\\ \Rightarrow \left\{ \begin{array}{l} a = 1\\ b = - 2\\ c = 2 \end{array} \right. \Rightarrow abc = 1.( - 2).2 = - 4 \end{array}\)

Câu hỏi này thuộc đề thi trắc nghiệm dưới đây, bấm vào Bắt đầu thi để làm toàn bài