Tính \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGMbGaai4jamaabmaabaGaaGymaaGaayjkaiaawMcaaaqaaiabeA8a % QjaacEcadaqadaqaaiaaicdaaiaawIcacaGLPaaaaaaaaa!3E89! \frac{{f'\left( 1 \right)}}{{\varphi '\left( 0 \right)}}\). Biết rằng : \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacI % cacaWG4bGaaiykaiabg2da9iaadIhadaahaaWcbeqaaiaaikdaaaaa % aa!3C21! f(x) = {x^2}\) và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeqOXdOMaai % ikaiaadIhacaGGPaGaeyypa0JaaGinaiaadIhacqGHRaWkciGGZbGa % aiyAaiaac6gadaWcaaqaaiabec8aWjaadIhaaeaacaaIYaaaaaaa!4408! \varphi (x) = 4x + \sin \frac{{\pi x}}{2}\).
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cacaGGOaGaamiEaiaacMcacqGH9aqpcaaIYaGaamiEaiabgkDiElaa % dAgacaGGNaGaaiikaiaaigdacaGGPaGaeyypa0JaaGOmaiaacUdacq % aHgpGAcaGGNaGaaiikaiaadIhacaGGPaGaeyypa0JaaGinaiabgUca % RmaalaaabaGaeqiWdahabaGaaGOmaaaaciGGJbGaai4Baiaacohada % Wcaaqaaiabec8aWjaadIhaaeaacaaIYaaaaiabgkDiElabeA8aQjaa % cEcacaGGOaGaaGimaiaacMcacqGH9aqpcaaI0aGaey4kaSYaaSaaae % aacqaHapaCaeaacaaIYaaaaaaa!6174! f'(x) = 2x \Rightarrow f'(1) = 2;\varphi '(x) = 4 + \frac{\pi }{2}\cos \frac{{\pi x}}{2} \Rightarrow \varphi '(0) = 4 + \frac{\pi }{2}\)
suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % WGMbGaai4jaiaacIcacaaIXaGaaiykaaqaaiabeA8aQjaacEcacaGG % OaGaaGimaiaacMcaaaGaeyypa0ZaaSaaaeaacaaI0aaabaGaaGioai % abgUcaRiabec8aWbaaaaa!435E! \frac{{f'(1)}}{{\varphi '(0)}} = \frac{4}{{8 + \pi }}\)