Đáp án:
Để pt có 2 nghiệm x1 và x2 thì:
$\begin{array}{l}
\Delta \ge 0\\
\Rightarrow {m^2} - 4.\left( { - 4} \right) \ge 0\\
\Rightarrow {m^2} + 16 \ge 0\left( {luon\,dung} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m\\
{x_1}{x_2} = - 4
\end{array} \right.\\
A = \dfrac{{2\left( {{x_1} + {x_2}} \right) + 7}}{{x_1^2 + x_2^2}}\\
= \dfrac{{2.m + 7}}{{{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}}}\\
= \dfrac{{2m + 7}}{{{m^2} - 2.\left( { - 4} \right)}}\\
= \dfrac{{2m + 7}}{{{m^2} + 8}}\\
\Rightarrow A.\left( {{m^2} + 8} \right) = 2m + 7\\
\Rightarrow A.{m^2} - 2m + 8A - 7 = 0\\
\Rightarrow \left\{ \begin{array}{l}
a = A\\
b = - 2\\
c = 8A - 7
\end{array} \right.\\
\Rightarrow \Delta ' \ge 0\\
\Rightarrow {\left( { - 1} \right)^2} - A.\left( {8A - 7} \right) \ge 0\\
\Rightarrow 1 - 8{A^2} + 7A \ge 0\\
\Rightarrow 8{A^2} - 7A - 1 \le 0\\
\Rightarrow \left( {8A + 1} \right)\left( {A - 1} \right) \le 0\\
\Rightarrow - \dfrac{1}{8} \le A \le 1\\
\Rightarrow \min A = - \dfrac{1}{8} \Leftrightarrow m = - 8
\end{array}$
