ĐK:\(\left[{}\begin{matrix}1\le x\le3\\x\le0\end{matrix}\right.\)
\(\sqrt{x^2-x}=\sqrt{3-x}\Leftrightarrow x^2-x=3-x\Leftrightarrow x^2=3\Leftrightarrow\)\(\left[{}\begin{matrix}x=\sqrt{3}\left(tm\right)\\x=-\sqrt{3}\left(tm\right)\end{matrix}\right.\)
Vậy S={\(\pm\sqrt{3}\)}