Trong các nghiệm ( x ; y ) thỏa mãn bất phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaadIhadaahaaadbeqaaiaaikdaaaWccqGH % RaWkcaaIYaGaamyEamaaCaaameqabaGaaGOmaaaaaSqabaGccaGGOa % GaaGOmaiaadIhacqGHRaWkcaWG5bGaaiykaiabgwMiZkaaigdaaaa!45EF! {\log _{{x^2} + 2{y^2}}}(2x + y) \ge 1\). Giá trị lớn nhất của biểu thức T = 2x + y bằng:
Hãy suy nghĩ và trả lời câu hỏi trước khi xem đáp án
Lời giải:
Báo saiBPT \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaci % iBaiaac+gacaGGNbWaaSbaaSqaaiaadIhadaahaaadbeqaaiaaikda % aaWccqGHRaWkcaaIYaGaamyEamaaCaaameqabaGaaGOmaaaaaSqaba % GccaGGOaGaaGOmaiaadIhacqGHRaWkcaWG5bGaaiykaiabgwMiZkaa % igdacqGHuhY2daGabaabaeqabaGaamiEamaaCaaaleqabaGaaGOmaa % aakiabgUcaRiaaikdacaWG5bWaaWbaaSqabeaacaaIYaaaaOGaeyOp % a4JaaGymaaqaaiaaikdacaWG4bGaey4kaSIaamyEaiabgwMiZkaadI % hadaahaaWcbeqaaiaaikdaaaGccqGHRaWkcaaIYaGaamyEamaaCaaa % leqabaGaaGOmaaaaaaGccaGL7baacaaMc8UaaGPaVlaaykW7caGGOa % GaamysaiaacMcacaGGSaGaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7 % caaMc8+aaiqaaqaabeqaaiaaicdacqGH8aapcaWG4bWaaWbaaSqabe % aacaaIYaaaaOGaey4kaSIaaGOmaiaadMhadaahaaWcbeqaaiaaikda % aaGccqGH8aapcaaIXaaabaGaaGimaiabgYda8iaaikdacaWG4bGaey % 4kaSIaamyEaiabgsMiJkaadIhadaahaaWcbeqaaiaaikdaaaGccqGH % RaWkcaaIYaGaamyEamaaCaaaleqabaGaaGOmaaaaaaGccaGL7baaca % aMc8UaaGPaVlaacIcacaWGjbGaamysaiaacMcaaaa!8B55! \Leftrightarrow {\log _{{x^2} + 2{y^2}}}(2x + y) \ge 1 \Leftrightarrow \left\{ \begin{array}{l} {x^2} + 2{y^2} > 1\\ 2x + y \ge {x^2} + 2{y^2} \end{array} \right.\,\,\,(I),\,\,\,\,\,\,\left\{ \begin{array}{l} 0 < {x^2} + 2{y^2} < 1\\ 0 < 2x + y \le {x^2} + 2{y^2} \end{array} \right.\,\,(II)\)
Xét T= 2x + y
TH1: (x; y) thỏa mãn (II) khi đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGimaiabgY % da8iaadsfacqGH9aqpcaaIYaGaamiEaiabgUcaRiaadMhacqGHKjYO % caWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGOmaiaadMhada % ahaaWcbeqaaiaaikdaaaGccqGH8aapcaaIXaaaaa!461C! 0 < T = 2x + y \le {x^2} + 2{y^2} < 1\)
TH2: (x; y) thỏa mãn (I) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaCa % aaleqabaGaaGOmaaaakiabgUcaRiaaikdacaWG5bWaaWbaaSqabeaa % caaIYaaaaOGaeyizImQaaGOmaiaadIhacqGHRaWkcaWG5bGaeyi1HS % TaaiikaiaadIhacqGHsislcaaIXaGaaiykamaaCaaaleqabaGaaGOm % aaaakiabgUcaRiaacIcadaGcaaqaaiaaikdaaSqabaGccaWG5bGaey % OeI0YaaSaaaeaacaaIXaaabaGaaGOmamaakaaabaGaaGOmaaWcbeaa % aaGccaGGPaWaaWbaaSqabeaacaaIYaaaaOGaeyizIm6aaSaaaeaaca % aI5aaabaGaaGioaaaaaaa!53B9! {x^2} + 2{y^2} \le 2x + y \Leftrightarrow {(x - 1)^2} + {(\sqrt 2 y - \frac{1}{{2\sqrt 2 }})^2} \le \frac{9}{8}\). Khi đó
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiaadI % hacqGHRaWkcaWG5bGaeyypa0JaaGOmaiaacIcacaWG4bGaeyOeI0Ia % aGymaiaacMcacqGHRaWkdaWcaaqaaiaaigdaaeaadaGcaaqaaiaaik % daaSqabaaaaOGaaiikamaakaaabaGaaGOmaaWcbeaakiaadMhacqGH % sisldaWcaaqaaiaaigdaaeaacaaIYaWaaOaaaeaacaaIYaaaleqaaa % aakiaacMcacqGHRaWkdaWcaaqaaiaaiMdaaeaacaaI0aaaaiabgsMi % JoaakaaabaGaaiikaiaaikdadaahaaWcbeqaaiaaikdaaaGccqGHRa % WkdaWcaaqaaiaaigdaaeaacaaIYaaaaiaacMcadaWadaqaaiaacIca % caWG4bGaeyOeI0IaaGymaiaacMcadaahaaWcbeqaaiaaikdaaaGccq % GHRaWkcaGGOaWaaOaaaeaacaaIYaaaleqaaOGaamyEaiabgkHiTmaa % laaabaGaaGymaaqaaiaaikdadaGcaaqaaiaaikdaaSqabaaaaOGaai % ykamaaCaaaleqabaGaaGOmaaaaaOGaay5waiaaw2faaaWcbeaakiab % gUcaRmaalaaabaGaaGyoaaqaaiaaisdaaaGaeyizIm6aaOaaaeaada % WcaaqaaiaaiMdaaeaacaaIYaaaaiaac6cadaWcaaqaaiaaiMdaaeaa % caaI4aaaaaWcbeaakiabgUcaRmaalaaabaGaaGyoaaqaaiaaisdaaa % Gaeyypa0ZaaSaaaeaacaaI5aaabaGaaGOmaaaaaaa!6E95! 2x + y = 2(x - 1) + \frac{1}{{\sqrt 2 }}(\sqrt 2 y - \frac{1}{{2\sqrt 2 }}) + \frac{9}{4} \le \sqrt {({2^2} + \frac{1}{2})\left[ {{{(x - 1)}^2} + {{(\sqrt 2 y - \frac{1}{{2\sqrt 2 }})}^2}} \right]} + \frac{9}{4} \le \sqrt {\frac{9}{2}.\frac{9}{8}} + \frac{9}{4} = \frac{9}{2}\)
Suy ra: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciyBaiaacg % gacaGG4bGaamivaiabg2da9maalaaabaGaaGyoaaqaaiaaikdaaaaa % aa!3C36! \max T = \frac{9}{2}\) \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaai % ikaiaadIhacaGG7aGaaGPaVlaacMhacaGGPaGaeyypa0Jaaiikaiaa % ikdacaGG7aGaaGPaVpaalaaabaGaaGymaaqaaiaaikdaaaGaaiykaa % aa!44D8! \Leftrightarrow (x;\,y) = (2;\,\frac{1}{2})\)