\(\Delta=1+4m>0\Rightarrow m>-\dfrac{1}{4}\)
Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=-1\\x_1x_2=-m\end{matrix}\right.\)
Do \(x_1\) là nghiệm nên: \(x_1^2+x_1-m=0\Rightarrow x_1^2=-x_1+m\)
\(\Rightarrow x_1^3=-x_1^2+mx_1=-\left(-x_1+m\right)+mx_1=x_1-m+mx_1\)
Ta được:
\(x_1^3+mx_2=-3\Leftrightarrow x_1-m+mx_1+mx_2=-3\)
\(\Leftrightarrow x_1-m+m\left(x_1+x_2\right)=-3\)
\(\Leftrightarrow x_1-m-m=-3\Rightarrow x_1=2m-3\)
\(\Rightarrow x_2=-1-x_1=-2m+2\)
Thế vào \(x_1x_2=-m\Rightarrow\left(2m-3\right)\left(-2m+2\right)=-m\)
\(\Leftrightarrow4m^2-11m+6=0\Rightarrow\left[{}\begin{matrix}m=\dfrac{3}{4}\\m=2\end{matrix}\right.\)
