Gọi M(a;b) là điểm thuộc đồ thị hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyEaiabg2 % da9maalaaabaGaaGOmaiaadIhacqGHRaWkcaaIXaaabaGaamiEaiab % gUcaRiaaikdaaaaaaa!3DF9! y = \frac{{2x + 1}}{{x + 2}}\) và có khoảng cách từ M đến đường thẳng \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizaiaacQ % dacaWG5bGaeyypa0JaaG4maiaadIhacqGHRaWkcaaI2aaaaa!3CFB! d:y = 3x + 6\) nhỏ nhất. Tìm giá trị của biểu thức \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 % da9iaaiodacaWGHbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamOy % amaaCaaaleqabaGaaGOmaaaaaaa!3D1B! T = 3{a^2} + {b^2}\).
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Lời giải:
Báo saiTa có \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamizamaabm % aabaGaamytaiaacUdacaaMc8UaamizaaGaayjkaiaawMcaaiabg2da % 9maalaaabaWaaqWaaeaacaaIZaGaamyyaiabgkHiTiaadkgacqGHRa % WkcaaI2aaacaGLhWUaayjcSdaabaWaaOaaaeaacaaIXaGaaGimaaWc % beaaaaaaaa!474C! d\left( {M;\,d} \right) = \frac{{\left| {3a - b + 6} \right|}}{{\sqrt {10} }}\) suy ra d(M;d) nhỏ nhất khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % aIZaGaamyyaiabgkHiTiaadkgacqGHRaWkcaaI2aaacaGLhWUaayjc % Sdaaaa!3E2F! \left| {3a - b + 6} \right|\) nhỏ nhất.
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % aIZaGaamyyaiabgkHiTiaadkgacqGHRaWkcaaI2aaacaGLhWUaayjc % SdGaeyypa0ZaaqWaaeaacaaIZaGaamyyaiabgkHiTmaalaaabaGaaG % OmaiaadggacqGHRaWkcaaIXaaabaGaamyyaiabgUcaRiaaikdaaaGa % ey4kaSIaaGOnaaGaay5bSlaawIa7aiabg2da9maaemaabaGaaG4mai % aadggacqGHRaWkcaaI0aGaey4kaSYaaSaaaeaacaaIZaaabaGaamyy % aiabgUcaRiaaikdaaaaacaGLhWUaayjcSdGaeyypa0ZaaqWaaeaaca % aIZaWaaeWaaeaacaWGHbGaey4kaSIaaGOmaaGaayjkaiaawMcaaiab % gUcaRmaalaaabaGaaG4maaqaaiaadggacqGHRaWkcaaIYaaaaiabgk % HiTiaaikdaaiaawEa7caGLiWoaaaa!66C8! \left| {3a - b + 6} \right| = \left| {3a - \frac{{2a + 1}}{{a + 2}} + 6} \right| = \left| {3a + 4 + \frac{3}{{a + 2}}} \right| = \left| {3\left( {a + 2} \right) + \frac{3}{{a + 2}} - 2} \right|\)
Nếu a > -2 thì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % aIZaWaaeWaaeaacaWGHbGaey4kaSIaaGOmaaGaayjkaiaawMcaaiab % gUcaRmaalaaabaGaaG4maaqaaiaadggacqGHRaWkcaaIYaaaaiabgk % HiTiaaikdaaiaawEa7caGLiWoacqGHLjYSdaabdaqaaiaaiAdacqGH % sislcaaIYaaacaGLhWUaayjcSdGaeyypa0JaaGinaaaa!4CD1! \left| {3\left( {a + 2} \right) + \frac{3}{{a + 2}} - 2} \right| \ge \left| {6 - 2} \right| = 4\).
Nếu a < -2 thì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % aIZaWaaeWaaeaacaWGHbGaey4kaSIaaGOmaaGaayjkaiaawMcaaiab % gUcaRmaalaaabaGaaG4maaqaaiaadggacqGHRaWkcaaIYaaaaiabgk % HiTiaaikdaaiaawEa7caGLiWoacqGH9aqpdaabdaqaaiabgkHiTiaa % iodadaqadaqaaiaadggacqGHRaWkcaaIYaaacaGLOaGaayzkaaGaey % 4kaSYaaSaaaeaacaaIZaaabaGaeyOeI0YaaeWaaeaacaWGHbGaey4k % aSIaaGOmaaGaayjkaiaawMcaaaaacqGHRaWkcaaIYaaacaGLhWUaay % jcSdGaeyyzImRaaGOnaiabgUcaRiaaikdacqGH9aqpcaaI4aaaaa!5BCE! \left| {3\left( {a + 2} \right) + \frac{3}{{a + 2}} - 2} \right| = \left| { - 3\left( {a + 2} \right) + \frac{3}{{ - \left( {a + 2} \right)}} + 2} \right| \ge 6 + 2 = 8\)
Vậy d(M;d) nhỏ nhất bằng 4 khi \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaaiaadggacqGH9aqpcqGHsislcaaIXaaabaGaamOyaiabg2da9iab % gkHiTiaaigdaaaGaay5Eaaaaaa!3E3E! \left\{ \begin{array}{l} a = - 1\\ b = - 1 \end{array} \right.\). Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 % da9iaaiodacaWGHbWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaamOy % amaaCaaaleqabaGaaGOmaaaakiabg2da9iaaisdaaaa!3EE9! T = 3{a^2} + {b^2} = 4\).