Biết rằng tồn tại duy nhất bộ các số nguyên \(a,b,c\) sao cho \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaa8qCaeaaca % GGOaGaaGinaiaadIhacqGHRaWkcaaIYaGaaiykaiGacYgacaGGUbGa % amiEaiaadsgacaWG4bGaeyypa0JaamyyaiabgUcaRiaadkgaciGGSb % GaaiOBaiaaikdacqGHRaWkcaWGJbGaciiBaiaac6gacaaIZaaaleaa % caaIYaaabaGaaG4maaqdcqGHRiI8aaaa!4E0E! \int\limits_2^3 {(4x + 2)\ln xdx = a + b\ln 2 + c\ln 3} \). Giá trị của \(a+b+c\) là
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Lời giải:
Báo saiXét \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9maapehabaGaaiikaiaaisdacaWG4bGaey4kaSIaaGOmaiaacMca % ciGGSbGaaiOBaiaadIhacaWGKbGaamiEaaWcbaGaaGOmaaqaaiaaik % daa0Gaey4kIipaaaa!4521! I = \int\limits_2^2 {(4x + 2)\ln xdx} \)
Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadwhacqGH9aqpciGGSbGaaiOBaiaadIhaaeaacaWGKbGaamOD % aiabg2da9iaacIcacaaI0aGaamiEaiabgUcaRiaaikdacaGGPaGaam % izaiaadIhaaaGaay5EaaGaeyO0H49aaiqaaqaabeqaaiaadsgacaWG % 1bGaeyypa0ZaaSaaaeaacaaIXaaabaGaamiEaaaacaWGKbGaamiEaa % qaaiaadAhacqGH9aqpcaaIYaGaamiEamaaCaaaleqabaGaaGOmaaaa % kiabgUcaRiaaikdacaWG4baaaiaawUhaaaaa!56D4! \left\{ \begin{array}{l} u = \ln x\\ dv = (4x + 2)dx \end{array} \right. \Rightarrow \left\{ \begin{array}{l} du = \frac{1}{x}dx\\ v = 2{x^2} + 2x \end{array} \right.\)
Khi đó
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamysaiabg2 % da9maaeiaabaWaamWaaeaadaqadaqaaiaaikdacaWG4bWaaWbaaSqa % beaacaaIYaaaaOGaey4kaSIaaGOmaiaadIhaaiaawIcacaGLPaaaci % GGSbGaaiOBaiaadIhaaiaawUfacaGLDbaaaiaawIa7amaaDaaaleaa % caaIYaaabaGaaG4maaaakiabgkHiTmaapehabaWaaeWaaeaacaaIYa % GaamiEaiabgUcaRiaaikdaaiaawIcacaGLPaaaaSqaaiaaikdaaeaa % caaIZaaaniabgUIiYdaaaa!505F! I = \left. {\left[ {\left( {2{x^2} + 2x} \right)\ln x} \right]} \right|_2^3 - \int\limits_2^3 {\left( {2x + 2} \right)} \)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacaaMb8 % UaaGzaVlaaygW7caaMb8UaaGPaVlaaykW7caaMc8Uaeyypa0JaaGOm % aiaaisdaciGGSbGaaiOBaiaaiodacqGHsislcaaIXaGaaGOmaiGacY % gacaGGUbGaaGOmaiabgkHiTmaaeiaabaWaaeWaaeaacaWG4bWaaWba % aSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiaadIhaaiaawIcacaGLPa % aaaiaawIa7amaaDaaaleaacaaIYaaabaGaaG4maaaaaOqaaiaaykW7 % caaMc8UaaGPaVlaaykW7cqGH9aqpcaaIYaGaaGinaiGacYgacaGGUb % GaaGinaiabgkHiTiaaigdacaaIYaGaciiBaiaac6gacaaIYaGaeyOe % I0IaaG4naaaaaa!672F! \begin{array}{l} \,\,\, = 24\ln 3 - 12\ln 2 - \left. {\left( {{x^2} + 2x} \right)} \right|_2^3\\ \,\,\,\, = 24\ln 4 - 12\ln 2 - 7 \end{array}\)
Vậy \(a=-7, b=-12, c=24\)
\(a=b+c=5\)
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
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