\(\Leftrightarrow mx-3-2x+2m=0\)
\(\Leftrightarrow m\left(x+2\right)-2\left(x+2\right)+1=0\)
\(\Leftrightarrow\left(m-2\right)\left(x+2\right)=-1\)
*Nếu x=2 PT trở thành 0.(x+2)=-1 ( vô nghiệm).Vậy x khác 2
Có \(x+2=-\frac{1}{m-2}\)
\(\Leftrightarrow x=-\frac{1}{m-2}-2\)
Để x nguyên thì \(-\frac{1}{m-2}\in Z\Rightarrow-1⋮m-2\Rightarrow m-2=\left(+-1\right)\Rightarrow m=\left(3,1\right)\)
(TMĐK)