\(\text{Δ}=\left[-\left(m+2\right)\right]^2-4\cdot1\cdot2m\)
\(=m^2+4m+4-8m=m^2-4m+4=\left(m-2\right)^2>=0\forall m\)
=>Phương trình luôn có nghiệm
Theo Vi-et, ta có:
\(\left\{{}\begin{matrix}x_1+x_2=-\dfrac{b}{a}=m+2\\x_1x_2=\dfrac{c}{a}=2m\end{matrix}\right.\)
\(-1< =\dfrac{2\left(x_1+x_2\right)}{x_1x_2}< =1\)
=>\(-1< =\dfrac{2\left(m+2\right)}{2m}< =1\)
=>\(-1< =\dfrac{m+2}{m}< =1\)
=>\(\left\{{}\begin{matrix}\dfrac{m+2}{m}+1>=0\\\dfrac{m+2}{m}-1< =0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{2m+2}{m}>=0\\\dfrac{2}{m}< =0\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}m< 0\\2m+2< =0\end{matrix}\right.\Leftrightarrow m< =-1\)
