Đáp án:
$\begin{array}{l}
a)m = 3\\
\Rightarrow {x^2} - x - 2 = 0\\
\Rightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = - 1
\end{array} \right.\\
b){x^2} - \left( {m - 2} \right)x + m - 5 = 0\\
\Delta = {\left( {m - 2} \right)^2} - 4m + 20\\
= {m^2} - 4m + 4 - 4m + 20\\
= {m^2} - 8m + 24\\
= {\left( {m - 4} \right)^2} + 8 > 0\\
THeo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m - 2\\
{x_1}{x_2} = m - 5
\end{array} \right.\\
A = 3\left( {x_1^2 + x_2^2} \right) + 8{x_1}{x_2}\\
= 3.{\left( {{x_1} + {x_2}} \right)^2} - 6{x_1}{x_2} + 8{x_1}{x_2}\\
= 3{\left( {{x_1} + {x_2}} \right)^2} + 2{x_1}{x_2}\\
= 3.{\left( {m - 2} \right)^2} + 2.\left( {m - 5} \right)\\
= 3{m^2} - 12m + 12 + 2m - 10\\
= 3{m^2} - 10m + 2\\
= 3\left( {{m^2} - 2.m.\dfrac{5}{3} + \dfrac{{25}}{9}} \right) - \dfrac{{25}}{3} + 2\\
= 3{\left( {m - \dfrac{5}{3}} \right)^2} - \dfrac{{19}}{3} \ge \dfrac{{ - 19}}{3}\\
\Rightarrow GTNN:A = - \dfrac{{19}}{3} \Leftrightarrow m = \dfrac{5}{3}
\end{array}$
