\(6-\left|2x-1\right|=x\)
\(\Leftrightarrow\left|2x-1\right|=6-x\)
\(\Leftrightarrow\left[{}\begin{matrix}2x-1=6-x\\2x-1=-6+x\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}3x=7\\x=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{7}{3}\\x=-5\end{matrix}\right.\)
Vậy \(S=\left\{-5;\dfrac{7}{3}\right\}\)