ta có : \(\left(m^2+1\right)x^2-\left(2-m\right)=0\Rightarrow2-m=\left(m^2+1\right)x^2\ge1\)
VẬY PT CÓ NGHIỆM KHI \(2-m\ge1\Leftrightarrow m\le1\).
\(\Rightarrow x^2=\frac{2-m}{m^2+1}\Leftrightarrow x=\sqrt{\frac{2-m}{m^2+1}}\)hoặc x=\(-\sqrt{\frac{2-m}{m^2+1}}\)