Ta có: \(\left|2x-1\right|-x=1\)
\(\Leftrightarrow\left|2x-1\right|=x+1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=x+1\left(x\ge\frac{1}{2}\right)\\1-2x=x+1\left(x< \frac{1}{2}\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\left(nhận\right)\\x=0\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2\right\}\)