\(a=m^2+m+1=\left(m+\dfrac{1}{2}\right)^2+\dfrac{3}{4}>0;\forall m\)
Phương trình có 2 nghiệm dương pb khi:
\(\left\{{}\begin{matrix}\Delta'=\left(2m-3\right)^2-\left(m+9\right)\left(m^2+m+1\right)>0\\x_1+x_2=\dfrac{-2\left(2m-3\right)}{m^2+m+1}>0\\x_1x_2=\dfrac{m+9}{m^2+m+1}>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-m^3-6m^2-22m>0\\-2\left(2m-3\right)>0\\m+9>0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}m\left[\left(m+3\right)^2+13\right]< 0\\2m-3< 0\\m+9>0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}m< 0\\m< \dfrac{3}{2}\\m>-9\end{matrix}\right.\) \(\Leftrightarrow-9< m< 0\)
