\(\frac{x+1}{-8}=\frac{-3}{x-1}\\ \Rightarrow\left(x+1\right)\left(x-1\right)=\left(-8\right)\left(-3\right)\\ \Rightarrow x\left(x-1\right)+\left(x-1\right)=24\\ \Rightarrow x^2-x+x-1=24\\ \Rightarrow x^2-1=24\\ \Rightarrow x^2=25\\ \Rightarrow\left[{}\begin{matrix}x=5\\x=-5\end{matrix}\right.\)
Vậy \(x\in\left\{5;-5\right\}\)