\(\frac{2x-m}{x-2}+\frac{x-1}{x+2}=3\)
\(\frac{\left(2x-m\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)}{x^2-4}=3\)(1)
\(đk:\left|x\right|\ne2\) \(\left(2x-m\right)\left(x+2\right)+\left(x-1\right)\left(x-2\right)=3\left(x^2-4\right)\)
\(2x^2+4x-mx-2m+\left(x^2-3x+2\right)=3x^2-12\)
\(\left(4-m-3\right)x=-12+2m-2\)
\(\left(1-m\right)x=2m-14\)
Với m=1 vô nghiệm
với \(m\ne0\Rightarrow x=\frac{2m-14}{1-m}\)
\(x>0\Rightarrow\frac{2\left(m-7\right)}{1-m}>0\Rightarrow1< m< 7\)(*)
\(\left|x\right|\ne2\Rightarrow\left|\frac{m-7}{1-m}\right|\ne1\) với (*)\(\Rightarrow\frac{m-7}{1-m}\ne1\Rightarrow m-7-\left(1-m\right)\ne0\Rightarrow2m\ne8\Rightarrow m\ne4\)
Kết luận: \(\left\{\begin{matrix}m\ne4\\1< m< 7\end{matrix}\right.\)