Đáp án:
$\begin{array}{l}
a)A = 4{x^2} - 10x\\
= {\left( {2x} \right)^2} - 2.2x.\dfrac{5}{2} + \dfrac{{25}}{4} - \dfrac{{25}}{4}\\
= {\left( {2x - \dfrac{5}{2}} \right)^2} - \dfrac{{25}}{4}\\
Do:{\left( {2x - \dfrac{5}{2}} \right)^2} \ge 0\\
\Leftrightarrow {\left( {2x - \dfrac{5}{2}} \right)^2} - \dfrac{{25}}{4} \ge \dfrac{{ - 25}}{4}\\
\Leftrightarrow A \ge \dfrac{{ - 25}}{4}\\
\Leftrightarrow GTNN:A = - \dfrac{{25}}{4}\\
Khi:x = \dfrac{5}{4}\\
Vậy\,x = \dfrac{5}{4}\\
b)B = {x^2} + 5{y^2} + 4xy + 6x - 2y + 2021\\
= {x^2} + 4{y^2} + 9 + 2.x.2y + 2.x.3 + 2.2y.3\\
+ {y^2} - 14y + 49 + 1963\\
= {\left( {x + 2y + 3} \right)^2} + {\left( {y - 7} \right)^2} + 1963\\
\left( {do:{{\left( {a + b + c} \right)}^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ac} \right)\\
\left\{ \begin{array}{l}
{\left( {x + 2y + 3} \right)^2} \ge 0\\
{\left( {y - 7} \right)^2} \ge 0
\end{array} \right.\\
\Leftrightarrow {\left( {x + 2y + 3} \right)^2} + {\left( {y - 7} \right)^2} + 1963 \ge 1963\\
\Leftrightarrow B \ge 1963\\
\Leftrightarrow GTNN:B = 1963\\
\Leftrightarrow \left\{ \begin{array}{l}
x + 2y + 3 = 0\\
y = 7
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = - 17\\
y = 7
\end{array} \right.\\
Vậy\,x = - 17;y = 7,GTNN:B = 1963
\end{array}$
