\(\frac{3x^2-6x+1}{x-2}=\frac{3x\left(x-2\right)+1}{x-2}=3x+\frac{1}{x-2}\)
Vì \(x\inℤ\)\(\Rightarrow3x\inℤ\)
\(\Rightarrow\)Để \(\frac{3x^2-6x+1}{x-2}\inℤ\)thì \(\frac{1}{x-2}\inℤ\)
\(\Rightarrow1⋮x-2\)\(\Rightarrow x-2\inƯ\left(1\right)=\left\{-1;1\right\}\)
\(\Rightarrow x\in\left\{1;3\right\}\)
Vậy \(x\in\left\{1;3\right\}\)