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