\(A=\frac{3x+2}{x-3}=\frac{3\left(x-3\right)+11}{x-3}=\frac{3\left(x-3\right)}{x-3}+\frac{11}{x-3}=3+\frac{11}{x-3}\left(ĐK:x\ne3\right)\)
Để A nguyên thì \(11⋮x-3\)hay \(x-3\inƯ\left(11\right)\)
Ư(11) | x - 3 | x |
1 | 1 | 4 |
-1 | -1 | 2 |
11 | 11 | 14 |
-11 | -11 | -8 |
Vậy để A nguyên \(x\in\left\{4;2;14;-8\right\}\)