Tính tích tất cả các nghiệm thực của phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaadaWcaaqaaiaa % ikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaGymaaqaai % aaikdacaWG4baaaaGaayjkaiaawMcaaiabgUcaRiaaikdadaahaaWc % beqaamaabmaabaGaamiEaiabgUcaRmaalaaabaGaaGymaaqaaiaaik % dacaWG4baaaaGaayjkaiaawMcaaaaakiabg2da9iaaiwdaaaa!4AD7! {\log _2}\left( {\frac{{2{x^2} + 1}}{{2x}}} \right) + {2^{\left( {x + \frac{1}{{2x}}} \right)}} = 5\)
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Lời giải:
Báo saiĐiều kiện: x > 0 .
PT: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyi1HSTaci % iBaiaac+gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOWaaeWaaeaadaWc % aaqaaiaaikdacaWG4bWaaWbaaSqabeaacaaIYaaaaOGaey4kaSIaaG % ymaaqaaiaaikdacaWG4baaaaGaayjkaiaawMcaaiabgUcaRiaaikda % daahaaWcbeqaamaabmaabaWaaSaaaeaacaaIYaGaamiEamaaCaaame % qabaGaaGOmaaaaliabgUcaRiaaigdaaeaacaaIYaGaamiEaaaaaiaa % wIcacaGLPaaaaaGccqGH9aqpcaaI1aGaaGPaVlaaykW7caaMc8UaaG % PaVlaaykW7daqadaqaaiaaigdaaiaawIcacaGLPaaaaaa!58DF! \Leftrightarrow {\log _2}\left( {\frac{{2{x^2} + 1}}{{2x}}} \right) + {2^{\left( {\frac{{2{x^2} + 1}}{{2x}}} \right)}} = 5\,\,\,\,\,\left( 1 \right)\)
Đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiDaiabg2 % da9maalaaabaGaaGOmaiaadIhadaahaaWcbeqaaiaaikdaaaGccqGH % RaWkcaaIXaaabaGaaGOmaiaadIhaaaGaeyypa0JaamiEaiabgUcaRm % aalaaabaGaaGymaaqaaiaaikdacaWG4baaaiabgwMiZkaaikdadaGc % aaqaaiaadIhacaGGUaWaaSaaaeaacaaIXaaabaGaaGOmaiaadIhaaa % aaleqaaOGaeyypa0ZaaOaaaeaacaaIYaaaleqaaaaa!4C25! t = \frac{{2{x^2} + 1}}{{2x}} = x + \frac{1}{{2x}} \ge 2\sqrt {x.\frac{1}{{2x}}} = \sqrt 2 \)
PT trở thành \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaciiBaiaac+ % gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOGaamiDaiabgUcaRiaaikda % daahaaWcbeqaaiaadshaaaGccqGH9aqpcaaI1aGaaeiiaiaabccaca % qGGaGaaeiiaiaabccacaqGGaGaaeiiaiaabccacaqGGaGaaeiiaiaa % bccacaqGGaGaaeiiaiaabccacaqGOaGaaeOmaiaabMcaaaa!4A38! {\log _2}t + {2^t} = 5{\rm { (2)}}\)
Xét hàm \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamOzamaabm % aabaGaamiDaaGaayjkaiaawMcaaiabg2da9iGacYgacaGGVbGaai4z % amaaBaaaleaacaaIYaaabeaakiaadshacqGHRaWkcaaIYaWaaWbaaS % qabeaacaWG0baaaOGaaGPaVlaaykW7daqadaqaaiaadshacqGHLjYS % daGcaaqaaiaaikdaaSqabaaakiaawIcacaGLPaaaaaa!4A2F! f\left( t \right) = {\log _2}t + {2^t}\,\,\left( {t \ge \sqrt 2 } \right)\) là hàm đồng biến nên:
\(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaeWaaeaaca % aIYaaacaGLOaGaayzkaaGaeyi1HSTaamOzamaabmaabaGaamiDaaGa % ayjkaiaawMcaaiabg2da9iaadAgadaqadaqaaiaaikdaaiaawIcaca % GLPaaacqGHuhY2caWG0bGaeyypa0JaaGOmaaaa!474F! \left( 2 \right) \Leftrightarrow f\left( t \right) = f\left( 2 \right) \Leftrightarrow t = 2\) (t/m).
Với t = 2 thì \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIYaGaamiEamaaCaaaleqabaGaaGOmaaaakiabgUcaRiaaigdaaeaa % caaIYaGaamiEaaaacqGH9aqpcaaIYaGaeyi1HSTaaGOmaiaadIhada % ahaaWcbeqaaiaaikdaaaGccqGHsislcaaI0aGaamiEaiabgUcaRiaa % igdacqGH9aqpcaaIWaaaaa!48D5! \frac{{2{x^2} + 1}}{{2x}} = 2 \Leftrightarrow 2{x^2} - 4x + 1 = 0\) (t/m). Vậy \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBa % aaleaacaaIXaaabeaakiaadIhadaWgaaWcbaGaaGOmaaqabaGccqGH % 9aqpdaWcaaqaaiaaigdaaeaacaaIYaaaaaaa!3C5E! {x_1}{x_2} = \frac{1}{2}\) (theo Viet ).
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
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