\(M=\dfrac{3x\left(x-3\right)+2}{x-3}\)
\(\Rightarrow M=\dfrac{3x\left(x-3\right)}{x-3}+\dfrac{2}{x-3}\)
\(\Rightarrow M=3x+\dfrac{2}{x-3}\)
Để M nguyên thì \(\dfrac{2}{x-3}\in Z\Rightarrow x-3\inƯ\left(2\right)=\left\{\pm1;\pm3\right\}\)
Ta có bảng:
x-3 | -1 | -3 | 1 | 3 |
x | 2 | 0 | 4 | 6 |
Vậy \(x\in\left\{2;0;4;6\right\}\)