Đáp án:
$\begin{array}{l}
a)m = 1\\
\Leftrightarrow {x^2} - 2x + 1 - 2 = 0\\
\Leftrightarrow {x^2} - 2x - 1 = 0\\
\Leftrightarrow {x^2} - 2x + 1 = 2\\
\Leftrightarrow {\left( {x - 1} \right)^2} = 2\\
\Leftrightarrow x = 1 \pm \sqrt 2 \\
Vậy\,x = 1 \pm \sqrt 2 \,khi:m = 1\\
b)\Delta ' \ge 0\\
\Leftrightarrow {m^2} - {m^2} + 2m \ge 0\\
\Leftrightarrow m \ge 0\\
Theo\,Viet:\left\{ \begin{array}{l}
{x_1} + {x_2} = 2m\\
{x_1}{x_2} = {m^2} - 2m
\end{array} \right.\\
Khi:P = {x_1}\left( {{x_1} + 1} \right) + {x_2}\left( {{x_2} + 1} \right) + 2021\\
= x_1^2 + x_2^2 + {x_1} + {x_2} + 2021\\
= {\left( {{x_1} + {x_2}} \right)^2} - 2{x_1}{x_2} + {x_1} + {x_2} + 2021\\
= {\left( {2m} \right)^2} - 2\left( {{m^2} - 2m} \right) + 2m + 2021\\
= 2{m^2} + 6m + 2021 \ge 2021\\
\Leftrightarrow P \ge 2021\\
\Leftrightarrow GTNN:P = 2021\\
Khi:m = 0
\end{array}$
