ĐKXĐ: x≥0
Ta có: \(\sqrt{x^2-2x+1}=2x\)
\(\Leftrightarrow\sqrt{\left(x-1\right)^2}=2x\)
\(\Leftrightarrow\left|x-1\right|=2x\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=2x\left(x\ge1\right)\\1-x=2x\left(x< 1\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1-2x=0\\1-x-2x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-1-x=0\\1-3x=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=1\\3x=1\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\left(loại\right)\\x=\frac{1}{3}\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x=\frac{1}{3}\)
