Ta có:
\(\dfrac{2x+1}{x-1}=\dfrac{2x-2+3}{x-1}=\dfrac{2\left(x-1\right)+3}{x-1}=2+\dfrac{3}{x-1}\)
Để \(2x+1\) chia hết cho x-1 thì:
\(x-1\in U\left(3\right)=\left\{1;-1;3;-3\right\}\)
Ta có bảng:
\(x-1\) | 1 | -1 | 3 | -3 |
x | 2 | 0 | 4 | -2 |
Vậy: \(x\in\left\{0;2;-2;4\right\}\)