\(\Delta'=\left(m+1\right)^2-\left(2m+1\right)=m^2\ge0;\forall m\) nên pt luôn có nghiệm
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=2m+1\end{matrix}\right.\)
\(\left(x_1+x_2\right)^2-\left(x_1x_2\right)^2-6m>4\)
\(\Leftrightarrow4\left(m+1\right)^2-\left(2m+1\right)^2-6m>4\)
\(\Leftrightarrow-2m>1\)
\(\Rightarrow m>-\dfrac{1}{2}\)