Có bao nhiêu số nguyên m để phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgU % caRiaaiodacqGH9aqpcaWGTbGaamyzamaaCaaaleqabaGaamiEaaaa % aaa!3C9C! x + 3 = m{e^x}\) có 2 nghiệm phân biệt?
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Lời giải:
Báo sai\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgU % caRiaaiodacqGH9aqpcaWGTbGaamyzamaaCaaaleqabaGaamiEaaaa % kiabgsDiBlaad2gacqGH9aqpdaWcaaqaaiaadIhacqGHRaWkcaaIZa % aabaGaamyzamaaCaaaleqabaGaamiEaaaaaaGccqGH9aqpcaWGMbWa % aeWaaeaacaWG4baacaGLOaGaayzkaaWaaeWaaeaacaGGQaaacaGLOa % GaayzkaaWaaeWaaeaacaWGebGaam4BaiaaykW7caaMc8UaaGPaVlaa % dwgadaahaaWcbeqaaiaadIhaaaGccqGH+aGpcaaIWaGaeyiaIiIaam % iEaiabgIGiolabl2riHcGaayjkaiaawMcaaaaa!5CF9! x + 3 = m{e^x} \Leftrightarrow m = \frac{{x + 3}}{{{e^x}}} = f\left( x \right)\left( * \right)\left( {Do\,\,\,{e^x} > 0\forall x \in R } \right)\)
Để phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgU % caRiaaiodacqGH9aqpcaWGTbGaamyzamaaCaaaleqabaGaamiEaaaa % aaa!3C9C! x + 3 = m{e^x}\) có 2 nghiệm phân biệt thì phương trình (*) có 2 nghiệm phân biệt.
Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiEaaGaayjkaiaawMcaaiabg2da9maalaaabaGaamiEaiab % gUcaRiaaiodaaeaacaWGLbWaaWbaaSqabeaacaWG4baaaaaaaaa!3F2B! f\left( x \right) = \frac{{x + 3}}{{{e^x}}}\) ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzaiaacE % cadaqadaqaaiaadIhaaiaawIcacaGLPaaacqGH9aqpdaWcaaqaaiaa % dwgadaahaaWcbeqaaiaadIhaaaGccqGHsisldaqadaqaaiaadIhacq % GHRaWkcaaIZaaacaGLOaGaayzkaaGaamyzamaaCaaaleqabaGaamiE % aaaaaOqaaiaadwgadaahaaWcbeqaaiaaikdacaWG4baaaaaakiabg2 % da9maalaaabaGaeyOeI0IaamiEaiabgkHiTiaaikdaaeaacaWGLbWa % aWbaaSqabeaacaWG4baaaaaakiabg2da9iaaicdacqGHuhY2caWG4b % Gaeyypa0JaeyOeI0IaaGOmaaaa!55DD! f'\left( x \right) = \frac{{{e^x} - \left( {x + 3} \right){e^x}}}{{{e^{2x}}}} = \frac{{ - x - 2}}{{{e^x}}} = 0 \Leftrightarrow x = - 2\)
Số nghiệm của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabg2 % da9iaadAgadaqadaqaaiaadIhaaiaawIcacaGLPaaaaaa!3B5D! m = f\left( x \right)\) là số giao điểm của đồ thị hàm số y = m và y =f(x)
Dựa vào BBT ta có phương trình (*) có 2 nghiệm phân biệt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4TaaG % imaiabgYda8iaad2gacqGH8aapcaWGLbWaaWbaaSqabeaacaaIYaaa % aaaa!3DD8! \Rightarrow 0 < m < {e^2}\)
Mà \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyBaiabgI % GiolablssiIkabgkDiElaad2gacqGHiiIZdaGadaqaaiaaigdacaGG % 7aGaaGOmaiaacUdacaaIZaGaai4oaiaaisdacaGG7aGaaGynaiaacU % dacaaI2aGaai4oaiaaiEdaaiaawUhacaGL9baaaaa!4A92! m \in Z \Rightarrow m \in \left\{ {1;2;3;4;5;6;7} \right\}\)
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
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