ĐKXĐ: \(x\ne\pm1\)
Ta có: \(A=\dfrac{x-1}{x^2-1}=\dfrac{1}{x+1}\)
Để \(A\in Z\Leftrightarrow\dfrac{1}{x+1}\in1\)
\(\Rightarrow x+1\inƯ_{\left(1\right)}\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=1\\x+1=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-2\end{matrix}\right.\) (tmđk)
Vậy để A nhận giá trị nguyên thì \(x=\left\{0;-2\right\}\)