\(A=\dfrac{8-x}{x-3}=\dfrac{-\left(x-3\right)+5}{x-3}=\dfrac{-\left(x-3\right)}{x-3}+\dfrac{5}{x-3}=-1+\dfrac{5}{x-3}\)
Để A thuộc Z thì \(\left(x-3\right)\inƯ\left(5\right)\)
Ta có \(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
Ta có bảng sau:
x-3 | -1 | 1 | -5 | 5 |
x | 2 | 4 | -2 | 8 |
Vậy \(x\in\left\{-2;2;4;8\right\}\)