Đáp án: $m = \dfrac{{16}}{7}$
Giải thích các bước giải:
$\begin{array}{l}
\Delta ' > 0\\
\Rightarrow {\left( {m - 4} \right)^2} + {m^2} + 2 > 0\left( {luôn\,đúng} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\left( {m - 4} \right) = 2m - 8\\
{x_1}{x_2} = - {m^2} - 2
\end{array} \right.\\
A = x_1^2 + x_2^2 - {x_1}{x_2}\\
= {\left( {{x_1} + {x_2}} \right)^2} - 3{x_1}{x_2}\\
= {\left( {2m - 8} \right)^2} - 3.\left( { - {m^2} - 2} \right)\\
= 4{m^2} - 32m + 64 + 3{m^2} + 6\\
= 7{m^2} - 32m + 70\\
= 7.\left( {{m^2} - 2.\dfrac{{16}}{7}.m + \dfrac{{256}}{{49}}} \right) + \dfrac{{234}}{7}\\
= 7.{\left( {m - \dfrac{{16}}{7}} \right)^2} + \dfrac{{234}}{7} \ge \dfrac{{234}}{7}\\
\Rightarrow GTNN:A = \dfrac{{234}}{7} \Leftrightarrow m = \dfrac{{16}}{7}
\end{array}$
