Mình còn cách khác :
Vì \(x-5⋮x-2\Rightarrow x-5-\left(x-2\right)⋮x-2\)
\(\Rightarrow x-5-x+2⋮x-2\)
\(\Rightarrow\left(x-x\right)-\left(5-2\right)⋮x-2\)
\(\Rightarrow0-3⋮x-2\)
\(\Rightarrow-3⋮x-2\)
\(\Rightarrow x-2\inƯ\left(-3\right)=\left\{1;-1;3;-3\right\}\)
| x - 2 | 1 | -1 | 3 | -3 |
| x | 3 | 1 | 5 | -1 |
Vậy \(x\in\left\{3;1;5;-1\right\}\)