Đáp án:
$\begin{array}{l}
{x^2} - \left( {m + 1} \right)x + m = 0\\
\Rightarrow \Delta > 0\\
\Rightarrow {\left( {m + 1} \right)^2} - 4m > 0\\
\Rightarrow {m^2} + 2m + 1 - 4m > 0\\
\Rightarrow {\left( {m - 1} \right)^2} > 0\\
\Rightarrow m \ne 1\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = m + 1\\
{x_1}{x_2} = m
\end{array} \right.\\
A = x_1^2{x_2} + {x_1}x_2^2 + 2020\\
= {x_1}{x_2}\left( {{x_1} + {x_2}} \right) + 2020\\
= m.\left( {m + 1} \right) + 2020\\
= {m^2} + m + 2020\\
= {m^2} + 2.m.\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{{8079}}{4}\\
= {\left( {m + \dfrac{1}{2}} \right)^2} + \dfrac{{8079}}{4} \ge \dfrac{{8079}}{4}\\
\Rightarrow GTNN:A = \dfrac{{8079}}{4}\\
Khi:m = - \dfrac{1}{2}
\end{array}$
