\(A=\frac{8-x}{x-3}=\frac{-\left(x-3\right)+5}{x-3}\)\(=\frac{-\left(x-3\right)}{x-3}+\frac{5}{x-3}\)\(=-1+\frac{5}{x-3}\)
Để \(A\in Z\) thì \(\left(x-3\right)\inƯ\left(5\right)\)
Ta có: \(Ư\left(5\right)=\left\{-1;1;-5;5\right\}\)
x-3 | -1 | 1 | -5 | 5 |
x | 2 | 4 | -2 | 8 |
Vậy \(x\in\left\{-2;2;4;8\right\}\)