Đáp án:
$\begin{array}{l}
B1)A = {x^2} + \left| {y + 3} \right| + 2021\\
Do:\left\{ \begin{array}{l}
{x^2} \ge 0\\
\left| {y + 3} \right| \ge 0
\end{array} \right.\\
\Rightarrow {x^2} + \left| {y + 3} \right| \ge 0\\
\Rightarrow {x^2} + \left| {y + 3} \right| + 2021 \ge 2021\\
\Rightarrow A \ge 2021\\
\Rightarrow GTNN:A = 2021\\
Khi:\left\{ \begin{array}{l}
x = 0\\
y = - 3
\end{array} \right.\\
B = {x^{2020}} + \left| {y + 1} \right| - 2019\\
Do:{x^{2020}} \ge 0;\left| {y + 1} \right| \ge 0\\
\Rightarrow {x^{2020}} + \left| {y + 1} \right| \ge 0\\
\Rightarrow {x^{2020}} + \left| {y + 1} \right| - 2019 \ge - 2019\\
\Rightarrow B \ge - 2019\\
\Rightarrow GTNN:B = - 2019\\
Khi:\left\{ \begin{array}{l}
x = 0\\
y = 1
\end{array} \right.
\end{array}$
