Đặt $x^2=y$
=> $y^2+my+(2m-4)=0$
Ta có
[tex]\Delta =m^2-4(2m-4)=m^2-8m+16=(m-4)^2[/tex]
[tex]\Leftrightarrow \left\{\begin{matrix} y_1=\frac{-m+\sqrt{\Delta }}{2}=\frac{-4}{2}=-2\\ y_2=\frac{-m-\sqrt{\Delta }}{2}=\frac{4-2m}{2}=2-m\\ \end{matrix}\right.[/tex]
=>[tex]\left\{\begin{matrix} x_1=\sqrt{2-m}\\ x_2=-\sqrt{2-m} \end{matrix}\right.[/tex]