Ta có: \(2x-\left|x+1\right|=\dfrac{-1}{2}\)
\(\Leftrightarrow\left|x+1\right|=2x+\dfrac{1}{2}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=2x+\dfrac{1}{2}\left(x\ge-1\right)\\x+1=-2x-\dfrac{1}{2}\left(x< -1\right)\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}-x=-\dfrac{1}{2}\\3x=-\dfrac{3}{2}\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\left(nhận\right)\\x=-\dfrac{1}{2}\left(loại\right)\end{matrix}\right.\)