Ta có: \(\frac{3-x}{-5}=\frac{5}{x-3}\)
\(\Leftrightarrow\frac{x-3}{5}=\frac{5}{x-3}\)
\(\Leftrightarrow\left(x-3\right)^2=25\)
\(\Leftrightarrow\left[{}\begin{matrix}x-3=5\\x-3=-5\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)(tm)
Vậy: x∈{-2;8}