a: \(\text{Δ}=\left[-\left(2m+1\right)\right]^2-4\left(m^2+m-6\right)\)
\(=4m^2+4m+1-4m^2-4m+24\)
=25>0
=>Phương trình luôn có hai nghiệm phân biệt
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=2m+1\\x_1x_2=\dfrac{c}{a}=m^2+m-6=\left(m+3\right)\left(m-2\right)\end{matrix}\right.\)
Để phương trình có hai nghiệm đều âm thì \(\left\{{}\begin{matrix}x_1+x_2< 0\\x_1x_2>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}2m+1< 0\\\left(m+3\right)\left(m-2\right)>0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< -\dfrac{1}{2}\\\left[{}\begin{matrix}m>2\\m< -3\end{matrix}\right.\end{matrix}\right.\Leftrightarrow m< -3\)
b: \(\left|x_1^3-x_2^3\right|=65\)
=>\(\left|\left(x_1-x_2\right)\left(x_1^2+x_1x_2+x_2^2\right)\right|=65\)
=>\(\left|\left(x_1-x_2\right)\cdot\left[\left(x_1+x_2\right)^2-x_1x_2\right]\right|=65\)
=>\(\left|\left(x_1-x_2\right)\left[\left(2m+1\right)^2-m^2-m+6\right]\right|=65\)
=>\(\left|\left(x_1-x_2\right)\left(4m^2+4m+1-m^2-m+6\right)\right|=65\)
=>\(\sqrt{\left(x_1-x_2\right)^2}\cdot\left|3m^2+3m+7\right|=65\)
=>\(\sqrt{\left(x_1+x_2\right)^2-4x_1x_2}\cdot\left|3m^2+3m+7\right|=65\)
=>\(\sqrt{\left(2m+1\right)^2-4\left(m^2+m-6\right)}\cdot\left|3m^2+3m+7\right|=65\)
=>\(5\cdot\left|3m^2+3m+7\right|=65\)
=>\(3m^2+3m+7=13\)
=>\(3m^2+3m-6=0\)
=>\(m^2+m-2=0\)
=>(m+2)(m-1)=0
=>\(\left[{}\begin{matrix}m=-2\\m=1\end{matrix}\right.\)