Đáp án:
$\begin{array}{l}
1)m = 2\\
\Rightarrow {x^2} + 9x - 4 = 0\\
\Rightarrow {x^2} + 2.x.\frac{9}{2} + \frac{{81}}{4} - \frac{{65}}{4} = 0\\
\Rightarrow {\left( {x + \frac{9}{2}} \right)^2} = \frac{{65}}{4}\\
\Rightarrow x = \frac{{ - 9 \pm \sqrt {65} }}{2}\\
2){x^2} + \left( {4m + 1} \right).x + 2\left( {m - 4} \right) = 0\\
\Rightarrow a.c < 0\\
\Rightarrow 1.2\left( {m - 4} \right) < 0\\
\Rightarrow m - 4 < 0\\
\Rightarrow m < 4\\
3)\Delta > 0\\
\Rightarrow {\left( {4m + 1} \right)^2} - 4.2.\left( {m - 4} \right) > 0\\
\Rightarrow 16{m^2} + 8m + 1 - 8m + 32 > 0\\
\Rightarrow 16{m^2} + 33 > 0\left( {tm} \right)\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = - 4m - 1\\
{x_1}{x_2} = 2\left( {m - 4} \right)
\end{array} \right.\\
a){x_1} - {x_2} = 17\\
\Rightarrow {x_1} = {x_2} + 17\\
\Rightarrow {x_2} + 17 + {x_2} = - 4m - 1\\
\Rightarrow {x_2} = - 2m - 9\\
\Rightarrow {x_1} = - 2m - 9 + 17\\
\Rightarrow {x_1} = - 2m + 8\\
\Rightarrow \left( { - 2m + 8} \right).\left( { - 2m - 9} \right) = 2\left( {m - 4} \right)\\
\Rightarrow \left( {m - 4} \right)\left( {2m + 9} \right) = m - 4\\
\Rightarrow \left( {m - 4} \right).\left( {2m + 9 - 1} \right) = 0\\
\Rightarrow \left( {m - 4} \right).\left( {2m + 8} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 4\\
m = - 4
\end{array} \right.\left( {tm} \right)\\
b)A = {\left( {{x_1} - {x_2}} \right)^2}\\
= {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2}\\
= {\left( {4m + 1} \right)^2} - 4.2.\left( {m - 4} \right)\\
= 16{m^2} + 8m + 1 - 8m + 32\\
= 16{m^2} + 33 \ge 33\\
\Rightarrow GTNN:A = 33\,khi:m = 0\\
c)\left\{ \begin{array}{l}
{x_1} + {x_2} = - 4m - 1\\
{x_1}{x_2} = 2\left( {m - 4} \right)
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
{x_1} + {x_2} = - 4m - 1\\
2{x_1}{x_2} = 4m - 16
\end{array} \right.\\
\Rightarrow {x_1} + {x_2} + 2{x_1}{x_2} = - 4m - 1 + 4m - 16\\
\Rightarrow {x_1} + {x_2} + 2{x_1}{x_2} = - 17
\end{array}$
