Giá trị của tham số m để phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGinamaaCa % aaleqabaGaamiEaaaakiabgkHiTiaad2gacaGGUaGaaGOmamaaCaaa % leqabaGaamiEaiabgUcaRiaaigdaaaGccqGHRaWkcaaIYaGaamyBai % abg2da9iaaicdaaaa!4254! {4^x} - m{.2^{x + 1}} + 2m = 0\) có hai nghiệm \(x_{1} ; x_{2}\), thoả mãn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEamaaBa % aaleaacaaIXaaabeaakiabgUcaRiaadIhadaWgaaWcbaGaaGOmaaqa % baGccqGH9aqpcaaIZaaaaa!3C76! {x_1} + {x_2} = 3\) là
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Lời giải:
Báo saiĐặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaW2dcaWG0b % Gaeyypa0JaaGOmamaaCaaaleqabaGaamiEaaaaaaa!3A56! t= {2^x}; t > 0\), . Phương trình trở thành: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaW2dcaWG0b % WaaWbaaSqabeaacaaIYaaaaOGaeyOeI0IaaGOmaiaad2gacaWG0bGa % ey4kaSIaaGOmaiaad2gacqGH9aqpcaaIWaaaaa!4041! {t^2} - 2mt + 2m = 0\) (1)
Phương trình đã cho có hai nghiệm \(x_{1} ; x_{2}\) , thỏa mãn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaW2dcaWG4b % WaaSbaaSqaaiaaigdaaeqaaOGaey4kaSIaamiEamaaBaaaleaacaaI % Yaaabeaakiabg2da9iaaiodaaaa!3CF3! {x_1} + {x_2} = 3\) khi và chỉ khi phương trình (1) có hai nghiệm dương phân biệt thỏa mãn \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaW2dcaWG0b % WaaSbaaSqaaiaaigdaaeqaaOGaaiOlaiaadshadaWgaaWcbaGaaGOm % aaqabaGccqGH9aqpcaaIYaWaaWbaaSqabeaacaWG4bWaaSbaaWqaai % aaigdaaeqaaaaakiaac6cacaaIYaWaaWbaaSqabeaacaWG4bWaaSba % aWqaaiaaikdaaeqaaaaakiabg2da9iaaikdadaahaaWcbeqaaiaadI % hadaWgaaadbaGaaGymaaqabaWccqGHRaWkcaWG4bWaaSbaaWqaaiaa % ikdaaeqaaaaakiabg2da9iaaikdadaahaaWcbeqaaiaaiodaaaGccq % GH9aqpcaaI4aaaaa!4D90! {t_1}.{t_2} = {2^{{x_1}}}{.2^{{x_2}}} = {2^{{x_1} + {x_2}}} = {2^3} = 8\)
Khi đó phương trình (1) có: \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbiqaaW2ddaGaba % abceqabaaBpeaacuqHuoargaqbaiabg2da9iaad2gadaahaaWcbeqa % aiaaikdaaaGccqGHsislcaaIYaGaamyBaiabg6da+iaaicdaaeaaca % WGtbGaeyypa0JaaGOmaiaad2gacqGH+aGpcaaIWaaabaGaamiuaiab % g2da9iaaikdacaWGTbGaeyOpa4JaaGimaaqaaiaadcfacqGH9aqpca % aIYaGaamyBaiabg2da9iaaiIdaaaGaay5EaaGaeyi1HSTaamyBaiab % g2da9iaaisdaaaa!55C8! \left\{ \begin{array}{l} \Delta ' = {m^2} - 2m > 0\\ S = 2m > 0\\ P = 2m > 0\\ P = 2m = 8 \end{array} \right. \Leftrightarrow m = 4\)
Đề thi thử tốt nghiệp THPT QG môn Toán năm 2020
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