\(ĐKXĐ:\left\{{}\begin{matrix}x-5\ge0\\1-x\ge0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge5\\x\le1\end{matrix}\right.\left(VL\right)}}\)
\(\Leftrightarrow2\sqrt{x-5}-\sqrt{x-5}=\sqrt{1-x}\)
=>căn (x-5)=căn (1-x)
=>x-5=1-x
=>2x=6
hay x=3(loại)