Cho hai số thực x,y thỏa mãn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaadIhadaahaaadbeqaaiaaikdaaaWccqGH % RaWkcaWG5bWaaWbaaWqabeaacaaIYaaaaSGaey4kaSIaaGOmaaqaba % GcdaqadaqaaiaadIhacqGHRaWkcaWG5bGaey4kaSIaaG4maaGaayjk % aiaawMcaaiabgwMiZkaaigdaaaa!47E5! {\log _{{x^2} + {y^2} + 2}}\left( {x + y + 3} \right) \ge 1\). Tìm giá trị lớn nhất của biểu thức S = 3x +4y -6.
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo saiTheo giải thiết ta có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabgU % caRiaadMhacqGHRaWkcaaIZaGaeyyzImRaamiEamaaCaaaleqabaGa % aGOmaaaakiabgUcaRiaadMhadaahaaWcbeqaaiaaikdaaaGccqGHRa % WkcaaIYaGaeyi1HS9aaeWaaeaacaWG4bGaeyOeI0YaaSaaaeaacaaI % XaaabaGaaGOmaaaaaiaawIcacaGLPaaadaahaaWcbeqaaiaaikdaaa % GccqGHRaWkdaqadaqaaiaadMhacqGHsisldaWcaaqaaiaaigdaaeaa % caaIYaaaaaGaayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgs % MiJoaalaaabaGaaG4maaqaaiaaikdaaaaaaa!54EE! x + y + 3 \ge {x^2} + {y^2} + 2 \Leftrightarrow {\left( {x - \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{2}} \right)^2} \le \frac{3}{2}\)
Khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaam4uaiabg2 % da9iaaiodadaqadaqaaiaadIhacqGHsisldaWcaaqaaiaaigdaaeaa % caaIYaaaaaGaayjkaiaawMcaaiabgUcaRiaaisdadaqadaqaaiaadM % hacqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaaaaGaayjkaiaawMca % aiabgkHiTmaalaaabaGaaGynaaqaaiaaikdaaaGaeyizIm6aaOaaae % aadaqadaqaaiaaiodadaahaaWcbeqaaiaaiodaaaGccqGHRaWkcaaI % 0aWaaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzkaaWaaeWaaeaada % qadaqaaiaadIhacqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaaaaGa % ayjkaiaawMcaamaaCaaaleqabaGaaGOmaaaakiabgUcaRmaabmaaba % GaamyEaiabgkHiTmaalaaabaGaaGymaaqaaiaaikdaaaaacaGLOaGa % ayzkaaWaaWbaaSqabeaacaaIYaaaaaGccaGLOaGaayzkaaaaleqaaO % GaeyOeI0YaaSaaaeaacaaI1aaabaGaaGOmaaaacqGHKjYOdaWcaaqa % aiaaiwdadaGcaaqaaiaaiAdaaSqabaGccqGHsislcaaI1aaabaGaaG % Omaaaaaaa!64D2! S = 3\left( {x - \frac{1}{2}} \right) + 4\left( {y - \frac{1}{2}} \right) - \frac{5}{2} \le \sqrt {\left( {{3^3} + {4^2}} \right)\left( {{{\left( {x - \frac{1}{2}} \right)}^2} + {{\left( {y - \frac{1}{2}} \right)}^2}} \right)} - \frac{5}{2} \le \frac{{5\sqrt 6 - 5}}{2}\)
Dấu bằng đạt tại \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaiqaaqaabe % qaamaalaaabaGaamiEaiabgkHiTmaalaaabaGaaGymaaqaaiaaikda % aaaabaGaaG4maaaacqGH9aqpdaWcaaqaaiaadMhacqGHsisldaWcaa % qaaiaaigdaaeaacaaIYaaaaaqaaiaaisdaaaGaaiilaaqaaiaaioda % caWG4bGaey4kaSIaaGinaiaadMhacqGHsislcaaIXaGaeyypa0ZaaS % aaaeaacaaI1aWaaOaaaeaacaaI2aaaleqaaOGaeyOeI0IaaGynaaqa % aiaaikdaaaaaaiaawUhaaiabgsDiBpaaceaaeaqabeaacaWG4bGaey % ypa0ZaaSaaaeaacaaIZaWaaOaaaeaacaaI2aaaleqaaOGaeyOeI0Ia % aGymaaqaaiaaigdacaaIWaaaaaqaaiaadMhacqGH9aqpdaWcaaqaai % aaisdadaGcaaqaaiaaiAdaaSqabaGccqGHsislcaaIZaaabaGaaGym % aiaaicdaaaaaaiaawUhaaaaa!5D90! \left\{ \begin{array}{l} \frac{{x - \frac{1}{2}}}{3} = \frac{{y - \frac{1}{2}}}{4},\\ 3x + 4y - 1 = \frac{{5\sqrt 6 - 5}}{2} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} x = \frac{{3\sqrt 6 - 1}}{{10}}\\ y = \frac{{4\sqrt 6 - 3}}{{10}} \end{array} \right.\)
