\(\Delta'=m^2-2m^2+1=1-m^2>0\Rightarrow-1< m< 1\) (1)
Để pt có 2 nghiệm đều dương:
\(\left\{{}\begin{matrix}x_1+x_2=2m>0\\x_1x_2=2m^2-1>0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}m>0\\\left[{}\begin{matrix}m>\frac{\sqrt{2}}{2}\\m< -\frac{\sqrt{2}}{2}\end{matrix}\right.\end{matrix}\right.\) \(\Rightarrow m>\frac{\sqrt{2}}{2}\)
Kết hợp (1) \(\Rightarrow\frac{\sqrt{2}}{2}< m< 1\)
\(x_1^3+x_2^3-\left(x_1^2+x_2^2\right)=-2\)
\(\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)-\left(x_1+x_2\right)^2+2x_1x_2+2=0\)
\(\Leftrightarrow8m^3-6m\left(2m^2-1\right)-4m^2+2\left(2m^2-1\right)=0\)
\(\Leftrightarrow2m^3-3m+1=0\)
\(\Leftrightarrow\left(m-1\right)\left(2m^2+2m-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}m=1\left(l\right)\\2m^2+2m-1=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}m=\frac{-1+\sqrt{3}}{2}\\m=\frac{-1-\sqrt{3}}{2}\left(l\right)\end{matrix}\right.\)
