Đáp án:
$\begin{array}{l}
a)m = 5\\
\Rightarrow 5{x^2} - 5x - 10 = 0\\
\Rightarrow {x^2} - x - 2 = 0\\
\Rightarrow \left( {x - 2} \right)\left( {x + 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 2\\
x = - 1
\end{array} \right.\\
b)\Delta = {\left( { - 5} \right)^2} + 4m.\left( {m + 5} \right)\\
= 25 + 4{m^2} + 20m\\
= 4\left( {{m^2} + 5m + \dfrac{{25}}{4}} \right)\\
= 4.{\left( {m + \dfrac{5}{2}} \right)^2} \ge 0\forall m
\end{array}$
=> phương trình luôn có nghiệm với mọi m
c) phương trình có 2 nghiệm phân biệt thì:
$\begin{array}{l}
\left\{ \begin{array}{l}
m \ne 0\\
\Delta > 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
m \ne 0\\
4.{\left( {m + \dfrac{5}{2}} \right)^2} > 0
\end{array} \right.\\
\Rightarrow m \ne - \dfrac{5}{2};m \ne 0\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = \dfrac{5}{m}\\
{x_1}{x_2} = - \dfrac{{m + 5}}{m}
\end{array} \right.\\
A = - 16{x_1}{x_2} - 3\left( {x_1^2 + x_2^2} \right)\\
= - 16.\left( { - \dfrac{{m + 5}}{m}} \right) - 3.\left( {{{\left( {{x_1} + {x_2}} \right)}^2} - 2{x_1}{x_2}} \right)\\
= 16.\dfrac{{m + 5}}{m} - 3.{\left( {\dfrac{5}{m}} \right)^2} + 6.\dfrac{{ - m - 5}}{m}\\
= 10.\dfrac{{m + 5}}{m} - \dfrac{{75}}{{{m^2}}}\\
A = 0\\
\Rightarrow 10.\dfrac{{m + 5}}{m} - \dfrac{{75}}{{{m^2}}} = 0\\
\Rightarrow 10.\left( {m + 5} \right).m - 75 = 0\\
\Rightarrow 2{m^2} + 10m - 15 = 0\\
\Rightarrow m = \dfrac{{ - 5 \pm \sqrt {55} }}{2}\left( {tmdk} \right)
\end{array}$
