\(\Delta=\left(m+2\right)^2-4\left(3m-3\right)=m^2-8m+16=\left(m-4\right)^2\)
TH1: \(\left\{{}\begin{matrix}m=4\\-\frac{b}{2a}=\frac{m+2}{2}< 5\end{matrix}\right.\) \(\Rightarrow m=4\)
TH2: \(m\ne4\) khi đó:
\(x_1< x_2< 5\Leftrightarrow\left\{{}\begin{matrix}\left(x_1-5\right)\left(x_2-5\right)>0\\\frac{x_1+x_2}{2}< 5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x_1x_2-5\left(x_1+x_2\right)+25>0\\x_1+x_2< 10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3m-3-5\left(m+2\right)+25>0\\m+2< 10\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}-2m+12>0\\m< 8\end{matrix}\right.\) \(\Rightarrow m< 6\)
\(\Rightarrow m=\left\{1;2;3;4;5\right\}\Rightarrow\sum m=15\)
