Cho hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaamiEaiabgUcaRiaaikdaaeaacaWG4bGaey4kaSIa % aGymaaaaaaa!3D3D! y = \frac{{x + 2}}{{x + 1}}\) có đồ thị là (C). Gọi d là khoảng cách từ giao điểm 2 tiệm cận của (C) đến một tiếp tuyến bất kỳ của (C). Giá trị lớn nhất có thể đạt được là:
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Lời giải:
Báo saiTiệm cận đứng là x = -1; tiệm cận ngang y = 1 nên I(-1;1) .
Gọi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytamaaBa % aaleaacaaIWaaabeaakmaabmaabaGaamiEamaaBaaaleaacaaIWaaa % beaakiaacUdacaqGGaWaaSaaaeaacaWG4bWaaSbaaSqaaiaaicdaae % qaaOGaey4kaSIaaGOmaaqaaiaadIhadaWgaaWcbaGaaGimaaqabaGc % cqGHRaWkcaaIXaaaaaGaayjkaiaawMcaaiabgIGiopaabmaabaGaam % 4qaaGaayjkaiaawMcaaaaa!4788! {M_0}\left( {{x_0};{\rm{ }}\frac{{{x_0} + 2}}{{{x_0} + 1}}} \right) \in \left( C \right)\);\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmOzayaafa % WaaeWaaeaacaWG4baacaGLOaGaayzkaaGaeyypa0JaeyOeI0YaaSaa % aeaacaaIXaaabaWaaeWaaeaacaWG4bGaey4kaSIaaGymaaGaayjkai % aawMcaamaaCaaaleqabaGaaGOmaaaaaaaaaa!413B! f'\left( x \right) = - \frac{1}{{{{\left( {x + 1} \right)}^2}}}\) nên phương trình tiếp tuyến của (C) là:
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabgk % HiTmaalaaabaGaamiEamaaBaaaleaacaaIWaaabeaakiabgUcaRiaa % ikdaaeaacaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSIaaGymaa % aacqGH9aqpcqGHsisldaWcaaqaaiaaigdaaeaadaqadaqaaiaadIha % daWgaaWcbaGaaGimaaqabaGccqGHRaWkcaaIXaaacaGLOaGaayzkaa % WaaWbaaSqabeaacaaIYaaaaaaakmaabmaabaGaamiEaiabgkHiTiaa % dIhadaWgaaWcbaGaaGimaaqabaaakiaawIcacaGLPaaacqGHuhY2da % WcaaqaaiaaigdaaeaadaqadaqaaiaadIhadaWgaaWcbaGaaGimaaqa % baGccqGHRaWkcaaIXaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYa % aaaaaakiaadIhacqGHRaWkcaWG5bGaeyOeI0YaaSaaaeaacaWG4bWa % a0baaSqaaiaaicdaaeaacaaIYaaaaOGaey4kaSIaaGinaiaadIhada % WgaaWcbaGaaGimaaqabaGccqGHRaWkcaaIYaaabaWaaeWaaeaacaWG % 4bWaaSbaaSqaaiaaicdaaeqaaOGaey4kaSIaaGymaaGaayjkaiaawM % caamaaCaaaleqabaGaaGOmaaaaaaGccqGH9aqpcaaIWaaaaa!69CA! y - \frac{{{x_0} + 2}}{{{x_0} + 1}} = - \frac{1}{{{{\left( {{x_0} + 1} \right)}^2}}}\left( {x - {x_0}} \right) \Leftrightarrow \frac{1}{{{{\left( {{x_0} + 1} \right)}^2}}}x + y - \frac{{x_0^2 + 4{x_0} + 2}}{{{{\left( {{x_0} + 1} \right)}^2}}} = 0\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamysaiaacYcacaqGGaGaeyiLdqeacaGLOaGaayzkaaGaeyyp % a0ZaaSaaaeaadaabdaqaaiabgkHiTmaalaaabaGaaGymaaqaamaabm % aabaGaamiEamaaBaaaleaacaaIWaaabeaakiabgUcaRiaaigdaaiaa % wIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaOGaey4kaSIaaGymai % abgkHiTmaalaaabaGaamiEamaaDaaaleaacaaIWaaabaGaaGOmaaaa % kiabgUcaRiaaisdacaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaey4kaS % IaaGOmaaqaamaabmaabaGaamiEamaaBaaaleaacaaIWaaabeaakiab % gUcaRiaaigdaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaaaaaa % GccaGLhWUaayjcSdaabaWaaOaaaeaadaWcaaqaaiaaigdaaeaadaqa % daqaaiaadIhadaWgaaWcbaGaaGimaaqabaGccqGHRaWkcaaIXaaaca % GLOaGaayzkaaWaaWbaaSqabeaacaaI0aaaaaaakiabgUcaRiaaigda % aSqabaaaaOGaeyypa0ZaaSaaaeaacaaIYaWaaqWaaeaacaWG4bWaaS % baaSqaaiaaicdaaeqaaOGaey4kaSIaaGymaaGaay5bSlaawIa7aaqa % amaakaaabaWaaeWaaeaacaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaey % 4kaSIaaGymaaGaayjkaiaawMcaamaaCaaaleqabaGaaGinaaaakiab % gUcaRiaaigdaaSqabaaaaOGaeyizImQaaGOmamaakaaabaWaaSaaae % aadaqadaqaaiaadIhadaWgaaWcbaGaaGimaaqabaGccqGHRaWkcaaI % XaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaGcbaGaaGOmam % aakaaabaWaaeWaaeaacaWG4bWaaSbaaSqaaiaaicdaaeqaaOGaey4k % aSIaaGymaaGaayjkaiaawMcaamaaCaaaleqabaGaaGinaaaaaeqaaa % aaaeqaaOGaeyypa0ZaaOaaaeaacaaIYaaaleqaaaaa!827E! d\left( {I,{\rm{ }}\Delta } \right) = \frac{{\left| { - \frac{1}{{{{\left( {{x_0} + 1} \right)}^2}}} + 1 - \frac{{x_0^2 + 4{x_0} + 2}}{{{{\left( {{x_0} + 1} \right)}^2}}}} \right|}}{{\sqrt {\frac{1}{{{{\left( {{x_0} + 1} \right)}^4}}} + 1} }} = \frac{{2\left| {{x_0} + 1} \right|}}{{\sqrt {{{\left( {{x_0} + 1} \right)}^4} + 1} }} \le 2\sqrt {\frac{{{{\left( {{x_0} + 1} \right)}^2}}}{{2\sqrt {{{\left( {{x_0} + 1} \right)}^4}} }}} = \sqrt 2 \)
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
Tuyển chọn số 3
