ĐKXĐ: \(x\ge\frac{4}{5}\)
\(pt\Leftrightarrow x^2-2x+1=5x-4+2\sqrt{5x-4}+1\)
\(\Leftrightarrow\left(x-1\right)^2=\left(\sqrt{5x-4}+1\right)^2\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{5x-4}+1=x-1\\\sqrt{5x-4}+1=1-x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}\sqrt{5x-4}=x-2\left(1\right)\\\sqrt{5x-4}=-x\left(2\right)\end{matrix}\right.\)
Vì \(x\ge\frac{4}{5}\Rightarrow\sqrt{5x-4}=-x\le-\frac{4}{5}\Rightarrow\) phương trình \(\left(2\right)\) vô nghiệm
\(\left(1\right)\Leftrightarrow\left\{{}\begin{matrix}5x-4=\left(x-2\right)^2\\x-2\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x^2-9x+8=0\\x\ge2\end{matrix}\right.\)
\(\Leftrightarrow x=8\left(tm\right)\)
