Cho \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaamiEaiabgUcaRiaaikdaaeaacaWG4bGaeyOeI0Ia % aGOmaaaacaaMc8UaaGPaVpaabmaabaGaam4qaaGaayjkaiaawMcaai % aac6caaaa!4361! y = \frac{{x + 2}}{{x - 2}}\,\,\left( C \right).\) Tìm M có hoành độ dương thuộc (C) sao cho tổng khoảng cách từ M đến 2 tiệm cận nhỏ nhất

Tập xác định: D = R\{2}

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGabmyEayaafa % Gaeyypa0ZaaSaaaeaacqGHsislcaaI0aaabaWaaeWaaeaacaWG4bGa % eyOeI0IaaGOmaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaa % GccaGGUaaaaa!3F92! y' = \frac{{ - 4}}{{{{\left( {x - 2} \right)}^2}}}.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamytaiabgI % GiopaabmaabaGaam4qaaGaayjkaiaawMcaaiabgkDiElaad2eadaqa % daqaaiaad2gacaGG7aWaaSaaaeaacaWGTbGaey4kaSIaaGOmaaqaai % aad2gacqGHsislcaaIYaaaaaGaayjkaiaawMcaaiaaykW7caaMc8+a % aeWaaeaacaWGTbGaeyOpa4JaaGimaaGaayjkaiaawMcaaaaa!4D91! M \in \left( C \right) \Rightarrow M\left( {m;\frac{{m + 2}}{{m - 2}}} \right)\,\,\left( {m > 0} \right)\)

Ta có 2 tiệm cận của (C) là: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaaBa % aaleaacaaIXaaabeaakiaacQdacaWG4bGaeyypa0JaaGOmaiaacUda % caaMc8UaaGPaVlaadsgadaWgaaWcbaGaaGOmaaqabaGccaGG6aGaam % yEaiabg2da9iaaigdacaGGUaaaaa!4529! {d_1}:x = 2;\,\,{d_2}:y = 1.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamyBaiaacYcacaWGKbWaaSbaaSqaaiaaigdaaeqaaaGccaGL % OaGaayzkaaGaey4kaSIaamizamaabmaabaGaamytaiaacYcacaWGKb % WaaSbaaSqaaiaaikdaaeqaaaGccaGLOaGaayzkaaGaeyypa0ZaaSaa % aeaadaabdaqaaiaad2gacqGHsislcaaIYaaacaGLhWUaayjcSdaaba % GaaGymaaaacqGHRaWkdaWcaaqaamaaemaabaWaaSaaaeaacaWGTbGa % ey4kaSIaaGOmaaqaaiaad2gacqGHsislcaaIYaaaaiabgkHiTiaaig % daaiaawEa7caGLiWoaaeaacaaIXaaaaiabg2da9maaemaabaGaamyB % aiabgkHiTiaaikdaaiaawEa7caGLiWoacqGHRaWkdaabdaqaamaala % aabaGaaGinaaqaaiaad2gacqGHsislcaaIYaaaaaGaay5bSlaawIa7 % aiaac6caaaa!64B4! d\left( {m,{d_1}} \right) + d\left( {M,{d_2}} \right) = \frac{{\left| {m - 2} \right|}}{1} + \frac{{\left| {\frac{{m + 2}}{{m - 2}} - 1} \right|}}{1} = \left| {m - 2} \right| + \left| {\frac{4}{{m - 2}}} \right|.\)

Áp dụng bất đẳng thức Cauchy cho 2 số dương |m -2|  và \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaada % WcaaqaaiaaisdaaeaacaWGTbGaeyOeI0IaaGOmaaaaaiaawEa7caGL % iWoaaaa!3C7E! \left| {\frac{4}{{m - 2}}} \right|\), ta có:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WGTbGaeyOeI0IaaGOmaaGaay5bSlaawIa7aiabgUcaRmaaemaabaWa % aSaaaeaacaaI0aaabaGaamyBaiabgkHiTiaaikdaaaaacaGLhWUaay % jcSdGaeyyzImRaaGOmamaakaaabaGaaGinaaWcbeaakiabg2da9iaa % isdaaaa!4846! \left| {m - 2} \right| + \left| {\frac{4}{{m - 2}}} \right| \ge 2\sqrt 4 = 4\)

Dấu “=” xảy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HS9aaq % WaaeaacaWGTbGaeyOeI0IaaGOmaaGaay5bSlaawIa7aiabg2da9maa % emaabaWaaSaaaeaacaaI0aaabaGaamyBaiabgkHiTiaaikdaaaaaca % GLhWUaayjcSdGaeyi1HS9aamqaaqaabeqaaiaad2gacqGHsislcaaI % YaGaeyypa0JaaGOmaaqaaiaad2gacqGHsislcaaIYaGaeyypa0Jaey % OeI0IaaGOmaaaacaGLBbaacqGHuhY2daWabaabaeqabaGaamyBaiab % g2da9iaaisdaaeaacaWGTbGaeyypa0JaaGimaaaacaGLBbaacqGHuh % Y2caWGTbGaeyypa0JaaGinaiaac6caaaa!612C! \Leftrightarrow \left| {m - 2} \right| = \left| {\frac{4}{{m - 2}}} \right| \Leftrightarrow \left[ \begin{array}{l} m - 2 = 2\\ m - 2 = - 2 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} m = 4\\ m = 0 \end{array} \right. \Leftrightarrow m = 4.\)

Vậy M(4;3)