\(\Delta'=\left(m+1\right)^2-m^2-2m=1\)
\(\Rightarrow\) phương trình đã cho luôn có 2 nghiệm pb
Theo Viet ta có: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=m^2+2m\end{matrix}\right.\)
Và do \(\Delta\) đẹp nên ta suy ra luôn \(\left|x_1-x_2\right|=\left|\frac{2\sqrt{\Delta'}}{a}\right|=2\)
\(\left|\left(x_1-x_2\right)\left(x_1^2+x_1x_2+x_2^2\right)\right|=8\)
\(\Leftrightarrow\left|x_1-x_2\right|.\left(x_1^2+x_1x_2+x_2^2\right)=8\) (do \(x_1^2+x_1x_2+x_2^2=\left(x_1+\frac{1}{2}x_2^2\right)+\frac{3x_2^2}{4}\ge0\))
\(\Leftrightarrow2\left(\left(x_1+x_2\right)^2-x_1x_2\right)=8\)
\(\Leftrightarrow4\left(m+1\right)^2-\left(m^2+2m\right)=4\)
\(\Leftrightarrow3m^2+6m=0\Rightarrow\left[{}\begin{matrix}m=0\\m=-\frac{1}{2}\end{matrix}\right.\)
