Question

Tìm x để \(\frac{\sqrt{x-3}}{\sqrt{2x+1}}=2\)?

Answer 1

\(\frac{\sqrt{x-3}}{\sqrt{2x+1}}=2\)

ĐKXĐ : \(x>3\)

=> \(\frac{\sqrt{x-3}}{\sqrt{2x+1}}.\sqrt{2x+1}=2\sqrt{2x+1}\)

=> \(\sqrt{x-3}=2\sqrt{2x+1}\)

=> \(\sqrt{\left(x-3\right)^2}=\left(2\sqrt{2x+1}\right)^2\)

=> \(\left|x-3\right|=2^2-\sqrt{\left(2x+1\right)^2}\)

=> \(\left[{}\begin{matrix}x-3=2^2-\sqrt{\left(2x+1\right)^2}\\x-3=-\left(2^2-\sqrt{\left(2x+1\right)^2}\right)\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}x-3=4-\left|2x+1\right|\\x-3=-4+\left|2x+1\right|\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}4-x+3=\left|2x+1\right|\\x-3+4=\left|2x+1\right|\end{matrix}\right.\)

=> \(\left[{}\begin{matrix}7-x=\left|2x+1\right|\\x+1=\left|2x+1\right|\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}7-x=2x+1\\7-x=-\left(2x+1\right)=-2x-1\end{matrix}\right.\\\left[{}\begin{matrix}x+1=2x+1\\x+1=-\left(2x+1\right)=-2x-1\end{matrix}\right.\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}7-1=x+2x\\7+1=x-2x\end{matrix}\right.\\\left[{}\begin{matrix}2x-x=1-1\\-2x-x=1+1\end{matrix}\right.\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}6=3x\\8=-x\end{matrix}\right.\\\left[{}\begin{matrix}x=0\\-3x=2\end{matrix}\right.\end{matrix}\right.\)

=> \(\left\{{}\begin{matrix}\left[{}\begin{matrix}x=6:3=2\left(loại\right)\\x=8:-1=-8\left(loại\right)\end{matrix}\right.\\\left[{}\begin{matrix}x=0\left(loại\right)\\x=2:\left(-3\right)=-\frac{2}{3}\left(loại\right)\end{matrix}\right.\end{matrix}\right.\)

Vậy \(x\in\varnothing\)

( đến cả máy tính casio ko tim đc nghiệm của bài toán ) :)

Answer 2

sửa hộ cái

ĐKXĐ : \(x\ge3\)

Answer 3

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaWaaSaaaeaada % GcaaqaaiaadIhacqGHsislcaaIZaaaleqaaaGcbaWaaOaaaeaacaaI % YaGaamiEaiabgUcaRiaaigdaaSqabaaaaOGaeyypa0JaaGOmaaaa!3E0B! \dfrac{{\sqrt {x - 3} }}{{\sqrt {2x + 1} }} = 2\). Điều kiện: \(x\ge3\)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGceaqabeaacqGHuh % Y2daGcaaqaaiaadIhacqGHsislcaaIZaaaleqaaOGaeyypa0JaaGOm % amaakaaabaGaaGOmaiaadIhacqGHRaWkcaaIXaaaleqaaaGcbaGaey % i1HSTaamiEaiabgkHiTiaaiodacqGH9aqpcaaI0aWaaeWaaeaacaaI % YaGaamiEaiabgUcaRiaaigdaaiaawIcacaGLPaaaaeaacqGHuhY2ca % WG4bGaeyOeI0IaaG4maiabg2da9iaaiIdacaWG4bGaey4kaSIaaGin % aaqaaiabgsDiBlaadIhacqGHsislcaaI4aGaamiEaiabg2da9iaais % dacqGHRaWkcaaIZaaabaGaeyi1HSTaeyOeI0IaamiEaiabg2da9iaa % iEdaaeaacqGHuhY2caWG4bGaeyypa0JaeyOeI0IaaG4namaabmaaba % Gaam4saiaadsfacaWGnbaacaGLOaGaayzkaaaaaaa!6EF6! \begin{array}{l} \Leftrightarrow \sqrt {x - 3} = 2\sqrt {2x + 1} \\ \Leftrightarrow x - 3 = 4\left( {2x + 1} \right)\\ \Leftrightarrow x - 3 = 8x + 4\\ \Leftrightarrow x - 8x = 4 + 3\\ \Leftrightarrow - x = 7\\ \Leftrightarrow x = - 7\left( {KTM} \right) \end{array}\)

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Answer 4

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGcbaWaaSaaaeaada % GcaaqaaiaadIhacqGHsislcaaIZaaaleqaaaGcbaWaaOaaaeaacaaI % YaGaamiEaiabgUcaRiaaigdaaSqabaaaaOGaeyypa0JaaGOmaaaa!3E0B! \dfrac{{\sqrt {x - 3} }}{{\sqrt {2x + 1} }} = 2\). Điều kiện: \(x\ge3\)

\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeaacaGaaiaabeqaamaabaabaaGceaqabeaacqGHuh % Y2daGcaaqaaiaadIhacqGHsislcaaIZaaaleqaaOGaeyypa0JaaGOm % amaakaaabaGaaGOmaiaadIhacqGHRaWkcaaIXaaaleqaaaGcbaGaey % i1HSTaamiEaiabgkHiTiaaiodacqGH9aqpcaaI0aWaaeWaaeaacaaI % YaGaamiEaiabgUcaRiaaigdaaiaawIcacaGLPaaaaeaacqGHuhY2ca % WG4bGaeyOeI0IaaG4maiabg2da9iaaiIdacaWG4bGaey4kaSIaaGin % aaqaaiabgsDiBlaadIhacqGHsislcaaI4aGaamiEaiabg2da9iaais % dacqGHRaWkcaaIZaaabaGaeyi1HSTaeyOeI0IaaG4naiaadIhacqGH % 9aqpcaaI3aaabaGaeyi1HSTaamiEaiabg2da9iabgkHiTiaaigdada % qadaqaaiaadUeacaWGubGaamytaaGaayjkaiaawMcaaaaaaa!6FB1! \begin{array}{l} \Leftrightarrow \sqrt {x - 3} = 2\sqrt {2x + 1} \\ \Leftrightarrow x - 3 = 4\left( {2x + 1} \right)\\ \Leftrightarrow x - 3 = 8x + 4\\ \Leftrightarrow x - 8x = 4 + 3\\ \Leftrightarrow - 7x = 7\\ \Leftrightarrow x = - 1\left( {KTM} \right) \end{array}\)

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