A,B là hai điểm di động và thuộc hai nhánh khác nhau của đồ thị \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaaGOmaiaadIhacqGHsislcaaIXaaabaGaamiEaiab % gUcaRiaaikdaaaaaaa!3E04! y = \frac{{2x - 1}}{{x + 2}}\). Khi đó khoảng cách AB bé nhất là?
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Lời giải:
Báo saiVì A, B thuộc hai nhánh của đồ thị \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaaGOmaiaadIhacqGHsislcaaIXaaabaGaamiEaiab % gUcaRiaaikdaaaaaaa!3E04! y = \frac{{2x - 1}}{{x + 2}}\) nên \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqamaabm % aabaGaamyyaiaacUdacaaIYaGaeyOeI0YaaSaaaeaacaaI1aaabaGa % amyyaiabgUcaRiaaikdaaaaacaGLOaGaayzkaaaaaa!3EE4! A\left( {a;2 - \frac{5}{{a + 2}}} \right)\) ; \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOqamaabm % aabaGaamOyaiaacUdacaaIYaGaeyOeI0YaaSaaaeaacaaI1aaabaGa % amOyaiabgUcaRiaaikdaaaaacaGLOaGaayzkaaaaaa!3EE7! B\left( {b;2 - \frac{5}{{b + 2}}} \right)\) với a > -2; b< -2
Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadk % eadaahaaWcbeqaaiaaikdaaaGccqGH9aqpdaqadaqaaiaadggacqGH % sislcaWGIbaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaai % OlamaadmaabaGaaGymaiabgUcaRmaalaaabaGaaGOmaiaaiwdaaeaa % daqadaqaaiaadggacqGHRaWkcaaIYaaacaGLOaGaayzkaaWaaWbaaS % qabeaacaaIYaaaaOWaaeWaaeaacaWGIbGaey4kaSIaaGOmaaGaayjk % aiaawMcaamaaCaaaleqabaGaaGOmaaaaaaaakiaawUfacaGLDbaacq % GH9aqpdaWadaqaamaabmaabaGaamyyaiabgUcaRiaaikdaaiaawIca % caGLPaaacqGHRaWkdaqadaqaaiabgkHiTiaadkgacqGHsislcaaIYa % aacaGLOaGaayzkaaaacaGLBbGaayzxaaWaaWbaaSqabeaacaaIYaaa % aOGaaiOlamaadmaabaGaaGymaiabgUcaRmaalaaabaGaaGOmaiaaiw % daaeaadaqadaqaaiaadggacqGHRaWkcaaIYaaacaGLOaGaayzkaaWa % aWbaaSqabeaacaaIYaaaaOWaaeWaaeaacqGHsislcaWGIbGaeyOeI0 % IaaGOmaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaaaaaakiaa % wUfacaGLDbaaaaa!6D21! A{B^2} = {\left( {a - b} \right)^2}.\left[ {1 + \frac{{25}}{{{{\left( {a + 2} \right)}^2}{{\left( {b + 2} \right)}^2}}}} \right] = {\left[ {\left( {a + 2} \right) + \left( { - b - 2} \right)} \right]^2}.\left[ {1 + \frac{{25}}{{{{\left( {a + 2} \right)}^2}{{\left( { - b - 2} \right)}^2}}}} \right]\)
Áp dụng bất đẳng thức Cô-si ta có:
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaamWaaeaada % qadaqaaiaadggacqGHRaWkcaaIYaaacaGLOaGaayzkaaGaey4kaSYa % aeWaaeaacqGHsislcaWGIbGaeyOeI0IaaGOmaaGaayjkaiaawMcaaa % Gaay5waiaaw2faamaaCaaaleqabaGaaGOmaaaakiabgwMiZkaaisda % daqadaqaaiaadggacqGHRaWkcaaIYaaacaGLOaGaayzkaaWaaeWaae % aacqGHsislcaWGIbGaeyOeI0IaaGOmaaGaayjkaiaawMcaaaaa!4E65! {\left[ {\left( {a + 2} \right) + \left( { - b - 2} \right)} \right]^2} \ge 4\left( {a + 2} \right)\left( { - b - 2} \right)(1)\)
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGymaiabgU % caRmaalaaabaGaaGOmaiaaiwdaaeaadaqadaqaaiaadggacqGHRaWk % caaIYaaacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaOGaaiOlam % aabmaabaGaeyOeI0IaamOyaiabgkHiTiaaikdaaiaawIcacaGLPaaa % daahaaWcbeqaaiaaikdaaaaaaOGaeyyzIm7aaSaaaeaacaaIXaGaaG % imaaqaamaabmaabaGaamyyaiabgUcaRiaaikdaaiaawIcacaGLPaaa % daqadaqaaiabgkHiTiaadkgacqGHsislcaaIYaaacaGLOaGaayzkaa % aaaaaa!5125! 1 + \frac{{25}}{{{{\left( {a + 2} \right)}^2}.{{\left( { - b - 2} \right)}^2}}} \ge \frac{{10}}{{\left( {a + 2} \right)\left( { - b - 2} \right)}}(2)\)
Từ (1) và (2) suy ra \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadk % eadaahaaWcbeqaaiaaikdaaaGccqGHLjYScaaI0aGaaGimaaaa!3BB2! A{B^2} \ge 40\)\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyO0H4Taam % yqaiaadkeacqGHLjYScaaIYaWaaOaaaeaacaaIXaGaaGimaaWcbeaa % aaa!3DF0! \Rightarrow AB \ge 2\sqrt {10} \)
Dấu "=" xảy ra khi và chỉ khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadggacqGHRaWkcaaIYaGaeyypa0JaeyOeI0IaaGOmaiabgkHi % TiaadkgaaeaacaaIXaGaeyypa0ZaaSaaaeaacaaIYaGaaGynaaqaam % aabmaabaGaamyyaiabgUcaRiaaikdaaiaawIcacaGLPaaadaahaaWc % beqaaiaaikdaaaGcdaqadaqaaiabgkHiTiaaikdacqGHsislcaWGIb % aacaGLOaGaayzkaaWaaWbaaSqabeaacaaIYaaaaaaaaaGccaGL7baa % cqGHuhY2daGabaabaeqabaGaamyyaiabg2da9maakaaabaGaaGynaa % WcbeaakiabgkHiTiaaikdaaeaacaWGIbGaeyypa0JaeyOeI0IaaGOm % aiabgkHiTmaakaaabaGaaGynaaWcbeaaaaGccaGL7baaaaa!59BE! \left\{ \begin{array}{l} a + 2 = - 2 - b\\ 1 = \frac{{25}}{{{{\left( {a + 2} \right)}^2}{{\left( { - 2 - b} \right)}^2}}} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} a = \sqrt 5 - 2\\ b = - 2 - \sqrt 5 \end{array} \right.\)
Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyqaiaadk % eadaWgaaWcbaGaciyBaiaacMgacaGGUbaabeaakiabg2da9iaaikda % daGcaaqaaiaaigdacaaIWaaaleqaaOGaaiOlaaaa!3E97! A{B_{\min }} = 2\sqrt {10} .\)