ĐKXĐ: \(x\ne1\)
Ta có: \(\dfrac{x-1}{-4}=\dfrac{-4}{x-1}\)
\(\Leftrightarrow\left(x-1\right)^2=16\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=4\\x-1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=5\left(nhận\right)\\x=-3\left(nhận\right)\end{matrix}\right.\)
Vậy: \(x\in\left\{5;-3\right\}\)