\(A=\frac{x^3-x^2+2}{x-1}\) ; ĐK: \(x\ne1\)
\(=\frac{x\left(x-1\right)+2}{x-1}\)\(=\frac{x\left(x-1\right)}{x-1}+\frac{2}{x-1}\)
\(=x+\frac{2}{x-1}\)
Mà: \(x\in Z\)
Để \(A\in Z\)\(\Leftrightarrow\)\(\frac{2}{x-1}\in Z\)
=> 2 chia hết cho x - 1
=> \(x-1\inƯ_{\left(2\right)}=\left\{\text{ }\pm1;\pm2\right\}\)
=> \(x\in\left\{2;0;3;-1\right\}\)(T/m)
Vậy \(x\in\left\{2;0;3;-1\right\}\).