Đáp án: m=2
Giải thích các bước giải:
$\begin{array}{l}
{x^2} - mx + m - 7 = 0\\
\Rightarrow \Delta > 0\\
\Rightarrow {m^2} - 4m + 28 > 0\\
\Rightarrow {m^2} - 4m + 4 + 24 > 0\\
\Rightarrow {\left( {m - 2} \right)^2} + 24 > 0\left( {luon\,dung} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m\\
{x_1}{x_2} = m - 7
\end{array} \right.\\
S = \dfrac{{2{x_1}{x_2} + 15}}{{x_1^2 + x_2^2 + 2\left( {{x_1}{x_2} + 1} \right)}}\\
= \dfrac{{2\left( {m - 7} \right) + 15}}{{{{\left( {{x_1} + {x_2}} \right)}^2} + 2}}\\
= \dfrac{{2\left( {m - 7} \right) + 15}}{{{m^2} + 2}}\\
\Rightarrow S.{m^2} + 2S = 2m + 1\\
\Rightarrow S.{m^2} - 2m + 2S - 1 = 0\\
\Rightarrow \Delta ' \ge 0\\
\Rightarrow 1 - S.\left( {2S - 1} \right) \ge 0\\
\Rightarrow 1 - 2{S^2} + S \ge 0\\
\Rightarrow 2{S^2} - S - 1 \le 0\\
\Rightarrow \left( {2S + 1} \right)\left( {S - 1} \right) \le 0\\
\Rightarrow - \dfrac{1}{2} \le S \le 1\\
\Rightarrow GTNN:S = - \dfrac{1}{2}\,khi:m = 2
\end{array}$
