\(\Delta'=\left(m+1\right)^2-2m-10=m^2-9\)
để pt có 2 no pb => \(\Delta'\ge0\Rightarrow m^2-9\ge0\Rightarrow\left\{{}\begin{matrix}m\ge3\\m\le-3\end{matrix}\right.\)
pt trên có no thì:
\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=2m+2\\x_1x_2=2m+1\end{matrix}\right.\)
mà
\(x_1^2+x_2^2+10x_1x_2=64\\ \Leftrightarrow\left(x_1+x_2\right)^2+6x_1x_2=64\\ \Leftrightarrow\left(2m+2\right)^2+6\left(2m+1\right)=64\\ \Leftrightarrow4m^2+20m+10-64=0\\ \Leftrightarrow.....\)
vậy m=...