Có bao nhiêu giá trị nguyên của tham số \(m\,\,(|m|<10)\)  để phương trình \(2^{x-1}=log_4{(x+2m)}+m\) có nghiệm

Điều kiện \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaaeaaaaaaaaa8 % qacaWG4bGaey4kaSIaaGOmaiaad2gacqGH+aGpcaaIWaaaaa!3B62! x + 2m > 0\)

Ta có:

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqababaaaaaaa % aapeqaaiaaikdadaahaaWcbeqaaiaadIhacqGHsislcaaIXaaaaOGa % eyypa0JaciiBaiaac+gacaGGNbWaaSbaaSqaaiaaisdaaeqaaOGaai % ikaiaadIhacqGHRaWkcaaIYaGaamyBaiaacMcacqGHRaWkcaWGTbaa % baGaeyi1HSTaaGOmamaaCaaaleqabaGaamiEaaaakiabg2da9iGacY % gacaGGVbGaai4zamaaBaaaleaacaaIYaaabeaakiaacIcacaWG4bGa % ey4kaSIaaGOmaiaad2gacaGGPaGaey4kaSIaaGOmaiaad2gaaaaa!55C0! \begin{array}{l} \,\,\,\,\,\,\,{2^{x - 1}} = {\log _4}(x + 2m) + m\\ \Leftrightarrow {2^x} = {\log _2}(x + 2m) + 2m \end{array}\)

đặt \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aacaWG0bGaeyypa0JaciiBaiaac+gacaGGNbWaaSbaaSqaaiaaikda % aeqaaOGaaiikaiaadIhacqGHRaWkcaaIYaGaamyBaiaacMcaaaa!40BA! t = {\log _2}(x + 2m)\)

khi đó có hệ phương trình: 

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aadaGabaabaeqabaGaaGOmamaaCaaaleqabaGaamiEaaaakiabg2da % 9iaadshacqGHRaWkcaaIYaGaamyBaaqaaiaadshacqGH9aqpciGGSb % Gaai4BaiaacEgadaWgaaWcbaGaaGOmaaqabaGccaGGOaGaamiEaiab % gUcaRiaaikdacaWGTbGaaiykaaaacaGL7baacqGHuhY2daGabaabae % qabaGaaGOmamaaCaaaleqabaGaamiEaaaakiabg2da9iaadshacqGH % RaWkcaaIYaGaamyBaaqaaiaaikdadaahaaWcbeqaaiaadshaaaGccq % GH9aqpcaWG4bGaey4kaSIaaGOmaiaad2gaaaGaay5EaaGaeyi1HS9a % aiqaaqaabeqaaiaaikdadaahaaWcbeqaaiaadIhaaaGccqGHsislca % WG0bGaeyypa0JaaGOmaiaad2gaaeaacaaIYaWaaWbaaSqabeaacaWG % 0baaaOGaeyOeI0IaamiEaiabg2da9iaaikdacaWGTbaaaiaawUhaaa % aa!6966! \left\{ \begin{array}{l} {2^x} = t + 2m\\ t = {\log _2}(x + 2m) \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {2^x} = t + 2m\\ {2^t} = x + 2m \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {2^x} - t = 2m\\ {2^t} - x = 2m \end{array} \right.\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aacqGHuhY2caaIYaWaaWbaaSqabeaacaWG4baaaOGaeyOeI0IaamiD % aiabg2da9iaaikdadaahaaWcbeqaaiaadshaaaGccqGHsislcaWG4b % Gaeyi1HSTaaGPaVlaaikdadaahaaWcbeqaaiaadIhaaaGccqGHRaWk % caWG4bGaeyypa0JaaGOmamaaCaaaleqabaGaamiDaaaakiabgUcaRi % aadshacaaMc8UaaGPaVlaaykW7caaMc8UaaGPaVlaaykW7caGGOaGa % aGymaiaacMcaaaa!58FA! \Leftrightarrow {2^x} - t = {2^t} - x \Leftrightarrow \,{2^x} + x = {2^t} + t\,\,\,\,\,\,(1)\)

Xét hàm số \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aacaWGMbGaaiikaiaadwhacaGGPaGaeyypa0JaaGOmamaaCaaaleqa % baGaamyDaaaakiabgUcaRiaadwhaaaa!3E20! f(u) = {2^u} + u\) là hàm số đồng biến trên \(\mathbb{R}\)

do đó \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aadaqadaqaaiaaigdaaiaawIcacaGLPaaacqGHuhY2caWG4bGaeyyp % a0JaamiDaaaa!3DAF! \left( 1 \right) \Leftrightarrow x = t\)

vậy ta có phương trình \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcqaaaaaaaaaWdbe % aacaaIYaWaaWbaaSqabeaacaWG4baaaOGaeyOeI0IaamiEaiabg2da % 9iaaikdacaWGTbaaaa!3CA1! {2^x} - x = 2m\)

Xét hàm số \(g(x)=2^x-x\)

\(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqababaaaaaaa % aapeqaaiaadEgacaGGNaGaaiikaiaadIhacaGGPaGaeyypa0JaaGOm % amaaCaaaleqabaGaamiEaaaakiGacYgacaGGUbGaaGOmaiabgkHiTi % aaigdaaeaacaWGNbGaai4jaiaacIcacaWG4bGaaiykaiabg2da9iaa % icdacqGHuhY2caaIYaWaaWbaaSqabeaacaWG4baaaOGaciiBaiaac6 % gacaaIYaGaeyOeI0IaaGymaiabg2da9iaaicdacqGHuhY2caWG4bGa % eyypa0JaciiBaiaac+gacaGGNbWaaSbaaSqaaiaaikdaaeqaaOWaae % WaaeaadaWcaaqaaiaaigdaaeaaciGGSbGaaiOBaiaaikdaaaaacaGL % OaGaayzkaaGaeyypa0JaeyOeI0IaciiBaiaac+gacaGGNbWaaSbaaS % qaaiaaikdaaeqaaOWaaeWaaeaaciGGSbGaaiOBaiaaikdaaiaawIca % caGLPaaaaaaa!6839! \begin{array}{l} g'(x) = {2^x}\ln 2 - 1\\ g'(x) = 0 \Leftrightarrow {2^x}\ln 2 - 1 = 0 \Leftrightarrow x = {\log _2}\left( {\frac{1}{{\ln 2}}} \right) = - {\log _2}\left( {\ln 2} \right) \end{array}\)

bảng biến thiên của hàm số \(g(x)\)

nhìn vào bảng biến thiên ta thấy phương trình (1) có nghiệm

 \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGceaqabeaacqGHuh % Y2caaIYaGaamyBaiabgwMiZoaalaaabaGaaGymaaqaaiGacYgacaGG % UbGaaGOmaaaacqGHRaWkciGGSbGaai4BaiaacEgadaWgaaWcbaGaaG % OmaaqabaGccaGGOaGaciiBaiaac6gacaaIYaGaaiykaaqaaiabgsDi % Blaad2gacqGHLjYSdaWcaaqaaiaaigdaaeaacaaIYaaaamaabmaaba % WaaSaaaeaacaaIXaaabaGaciiBaiaac6gacaaIYaaaaiabgUcaRiGa % cYgacaGGVbGaai4zamaaBaaaleaacaaIYaaabeaakiaacIcaciGGSb % GaaiOBaiaaikdacaGGPaaacaGLOaGaayzkaaGaeyisISRaaGimaiaa % cYcacaaI1aaaaaa!5FD8! \begin{array}{l} \Leftrightarrow 2m \ge \frac{1}{{\ln 2}} + {\log _2}(\ln 2)\\ \Leftrightarrow m \ge \frac{1}{2}\left( {\frac{1}{{\ln 2}} + {{\log }_2}(\ln 2)} \right) \approx 0,5 \end{array}\)

Kết hợp với điều kiện \(% MathType!MTEF!2!1!+- % feaahqart1ev3aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaqWaaeaaca % WGTbaacaGLhWUaayjcSdGaeyipaWJaaGymaiaaicdaaaa!3C80! \left| m \right| < 10\) thì có 9 giá trị m để phương trình có nghiệm