Đáp án:
$\begin{array}{l}
\left( {{m^2} + m + 1} \right).{x^2} - \left( {{m^2} + 8m + 3} \right).x - 1 = 0\\
\Leftrightarrow \left\{ \begin{array}{l}
a = {m^2} + m + 1\\
b = - \left( {{m^2} + 8m + 3} \right)\\
c = - 1
\end{array} \right.\\
Theo\,Viet:{x_1}.{x_2} = \dfrac{c}{a} = \dfrac{{ - 1}}{{\left( {{m^2} + m + 1} \right)}}\\
Do:{m^2} + m + 1\\
= {m^2} + 2.m.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}\\
= {\left( {m + \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4} > 0\\
\Leftrightarrow \dfrac{1}{{{m^2} + m + 1}} > 0\\
\Leftrightarrow \dfrac{{ - 1}}{{{m^2} + m + 1}} < 0\\
\Leftrightarrow {x_1}.{x_2} < 0\\
S = {x_1} + {x_2} = \dfrac{{ - b}}{a} = \dfrac{{{m^2} + 8m + 3}}{{{m^2} + m + 1}}\\
\Leftrightarrow S.{m^2} + S.m + S = {m^2} + 8m + 3\\
\Leftrightarrow \left( {S - 1} \right).{m^2} + \left( {S - 8} \right).m + S - 3 = 0\\
\Leftrightarrow \Delta \ge 0\\
\Leftrightarrow {\left( {S - 8} \right)^2} - 4\left( {S - 1} \right)\left( {S - 3} \right) \ge 0\\
\Leftrightarrow {S^2} - 16S + 64 - 4{S^2} - 16S + 12 \ge 0\\
\Leftrightarrow 3{S^2} + 32S - 76 \le 0\\
\Leftrightarrow \left( {3S + 38} \right)\left( {S - 2} \right) \le 0\\
\Leftrightarrow \dfrac{{ - 38}}{3} \le S \le 2\\
\Leftrightarrow \left\{ \begin{array}{l}
GTNN:S = - \dfrac{{38}}{3}\\
GTLN:S = 2
\end{array} \right.
\end{array}$
