Ta có: \(\left|2x-1\right|=\left|x+2\right|\)
\(\Rightarrow\left[{}\begin{matrix}2x-1=x+2\\2x-1=-x-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}2x-x=1+2\\2x+x=1-2\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=3\\3x=-1\Rightarrow x=\dfrac{-1}{3}\end{matrix}\right.\)
Vậy \(\left[{}\begin{matrix}x=3\\x=\dfrac{-1}{3}\end{matrix}\right.\).