\(\Delta=\left(m+1\right)^2-4\left(m-2\right)=m^2-2m+9=\left(m+1\right)^2+8>0\)
Vậy pt luôn có 2 nghiệm pb
Theo Vi et \(\left\{{}\begin{matrix}x_1+x_2=m+1\\x_1x_2=m-2\end{matrix}\right.\)
Ta có \(\left(x_1+x_2\right)^2+3x_1x_2=-3\)
\(\left(m+1\right)^2+3\left(m-2\right)=-3\)
\(\Leftrightarrow m^2+5m-5=-3\Leftrightarrow m=\dfrac{-5\pm\sqrt{33}}{2}\)