*hoặc không =))
\(Q=\dfrac{x+3}{x-3}=1+\dfrac{6}{x-3}\) (x \(\ne\) 3)
Để \(Q\in Z\Rightarrow\left\{{}\begin{matrix}1\in Z\\\dfrac{6}{x-3}\in Z\end{matrix}\right.\)
Để \(\dfrac{6}{x-3}\in Z\Rightarrow x-3\inƯ\left(6\right)=\left\{\pm1;\pm2;\pm3;\pm6\right\}\)
| x-3 | 1 | -1 | 2 | -2 | 3 | -3 | 6 | -6 |
| x | 4 (TM) | 2 (TM) | 5 (TM) | 1 (TM) | 6 (TM) | 0 (TM) | 9 (TM) | -3 (TM) |
\(\Rightarrow x\in\left\{4;2;5;1;6;0;9;-3\right\}\) thì Q \(\in Z\)