ĐK: \(1\le x\le5\)
\(\sqrt{5-x}-\sqrt{x-1}=2\Leftrightarrow\left(\sqrt{5-x}-\sqrt{x-1}\right)=2^2\Leftrightarrow5-x+x-1-2\sqrt{\left(5-x\right)\left(x-1\right)}=4\Leftrightarrow2\sqrt{\left(5-x\right)\left(x-1\right)}=0\Leftrightarrow\sqrt{\left(5-x\right)\left(x-1\right)}=0\Leftrightarrow\left(5-x\right)\left(x-1\right)=0\Leftrightarrow\)\(\left[{}\begin{matrix}5-x=0\\x-1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left[{}\begin{matrix}x=5\left(tm\right)\\x=1\left(tm\right)\end{matrix}\right.\)
Vậy S={1;5}