Đáp án:
$\begin{array}{l}
\Delta ' = {m^2} - m + 2\\
= {\left( {m - \dfrac{1}{2}} \right)^2} + \dfrac{7}{4} > 0
\end{array}$
=> pt có 2 nghiệm phân biệt với mọi m
$\begin{array}{l}
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m\\
{x_1}{x_2} = m - 2
\end{array} \right.\\
x_1^2 + x_2^2 - 6{x_1}{x_2}\\
= {\left( {{x_1} + {x_2}} \right)^2} - 8{x_1}{x_2}\\
= 4{m^2} - 8.\left( {m - 2} \right)\\
= 4{m^2} - 8m + 16\\
= 4\left( {{m^2} - 2m + 1} \right) + 12\\
= 4{\left( {m - 1} \right)^2} + 12 \ge 12\\
\Leftrightarrow \dfrac{1}{{x_1^2 + x_2^2 - 6{x_1}{x_2}}} \le \dfrac{1}{{12}}\\
\Leftrightarrow \dfrac{{ - 24}}{{x_1^2 + x_2^2 - 6{x_1}{x_2}}} \ge \dfrac{{ - 24}}{{12}}\\
\Leftrightarrow M \ge - 2\\
\Leftrightarrow GTNN:M = - 2\\
Khi:m = 1\\
Vậy\,m = 1
\end{array}$
