Đáp án:
Bài 1:
Để pt có 2 nghiệm thì:
$\begin{array}{l}
\Delta ' \ge 0\\
\Rightarrow {\left( { - 1} \right)^2} - m + 3 \ge 0\\
\Rightarrow m \le 4\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2\\
{x_1}{x_2} = m - 3
\end{array} \right.\\
H = x_1^2x_2^2 - {x_1}{x_2} + 2019\\
= {\left( {{x_1}{x_2}} \right)^2} - {x_1}{x_2} + 2019\\
= {\left( {m - 3} \right)^2} - m + 3 + 2019\\
= {m^2} - 7m + 2031\\
B2:\\
a)\Delta ' \ge 0\\
\Rightarrow {\left( {m + 1} \right)^2} - {m^2} + 2m + 5 \ge 0\\
\Rightarrow 4m + 6 \ge 0\\
\Rightarrow m \ge - \dfrac{3}{2}\\
b)Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = - 2\left( {m + 1} \right)\\
{x_1}{x_2} = {m^2} - 2m - 5
\end{array} \right.\\
3{x_1} + 3{x_2} = - \dfrac{1}{2}{x_1}{x_2}\\
\Rightarrow 3.\left( { - 2m - 2} \right) = - \dfrac{1}{2}.\left( {{m^2} - 2m - 5} \right)\\
\Rightarrow 12m + 12 = {m^2} - 2m - 5\\
\Rightarrow {m^2} - 14m - 17 = 0\\
\Rightarrow \left[ \begin{array}{l}
m = 7 + \sqrt {66} \left( {tm} \right)\\
m = 7 - \sqrt {66} \left( {tm} \right)
\end{array} \right.
\end{array}$
