đặt m+1=t
\(\Leftrightarrow x^2-4tx+t=0\Leftrightarrow x^2-4tx+4t^2+t-4t^2=0\)
\(\Leftrightarrow\left(x-2t\right)^2=4t^2-t\)
để 2 có nghiệm \(\Rightarrow4t^2-t>0\) \(\Leftrightarrow t\left(4t-1\right)>0\Rightarrow\orbr{\begin{cases}t< 0\\t>\frac{1}{4}\end{cases}}\Rightarrow\hept{\begin{cases}m< -1\\m>-\frac{3}{4}\end{cases}}\)