Để PT có 2 nghiệm \(\Leftrightarrow\Delta=\left(m-1\right)^2-4\left(m+6\right)\ge0\)
\(\Leftrightarrow m^2-6m-23\ge0\\ \Leftrightarrow\left[{}\begin{matrix}m\le3-4\sqrt{2}\\m\ge3+4\sqrt{2}\end{matrix}\right.\)
Áp dụng Viét: \(\left\{{}\begin{matrix}x_1+x_2=1-m\\x_1x_2=m+6\end{matrix}\right.\)
\(x_1^2+x_2^2=10\\ \Leftrightarrow\left(x_1+x_2\right)^2-2x_1x_2=10\\ \Leftrightarrow\left(1-m\right)^2-2\left(m+6\right)=10\\ \Leftrightarrow m^2-2m+1-2m-12=10\\ \Leftrightarrow m^2-4m-21=0\\ \Leftrightarrow\left[{}\begin{matrix}m=7\left(ktm\right)\\m=-3\left(tm\right)\end{matrix}\right.\Leftrightarrow m=-3\)