Ta có: \(\left|2x+1\right|-3x+1=2\)
\(\Leftrightarrow\left|2x+1\right|=2+3x-1\)
\(\Leftrightarrow\left|2x+1\right|=3x+1\)
\(\Leftrightarrow\left[{}\begin{matrix}2x+1=3x+1\left(x\ge-\dfrac{1}{2}\right)\\-2x-1=3x+1\left(x< \dfrac{-1}{2}\right)\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=-\dfrac{2}{5}\left(loại\right)\end{matrix}\right.\)
Vậy: S={0}