Đáp án:
$\begin{array}{l}
a)A = {x^2} + 4{y^2} - 2xy - 6y - 10\left( {x - y} \right) + 32\\
= {x^2} - 2xy + {y^2} - 10\left( {x - y} \right) + 3{y^2} - 6y + 32\\
= {\left( {x - y} \right)^2} - 10\left( {x - y} \right) + 3\left( {{y^2} - 2y + 1} \right) + 29\\
= {\left( {x - y} \right)^2} - 10\left( {x - y} \right) + 25 + 3{\left( {y - 1} \right)^2} + 4\\
= {\left( {x - y - 5} \right)^2} + 3{\left( {y - 1} \right)^2} + 4\\
Do:\left\{ \begin{array}{l}
{\left( {x - y - 5} \right)^2} \ge 0\\
3{\left( {y - 1} \right)^2} \ge 0
\end{array} \right.\\
\Rightarrow A \ge 4\\
\Rightarrow GTNN:A = 4\\
Khi:\left\{ \begin{array}{l}
x - y - 5 = 0\\
y - 1 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
x = 6\\
y = 1
\end{array} \right.\\
b){y^2} - 2xy + 3{x^2} + 2y - 14x + 2020\\
= {y^2} + {x^2} + 1 - 2xy + 2y - 2x + 2{x^2} - 12x + 2019\\
= {\left( {y - x + 1} \right)^2} + 2\left( {{x^2} - 6x + 9} \right) + 2001\\
= {\left( {y - x + 1} \right)^2} + 2{\left( {x - 3} \right)^2} + 2001\\
\ge 2001\\
\Rightarrow GTNN = 2001\\
Khi:\left\{ \begin{array}{l}
y - x + 1 = 0\\
x - 3 = 0
\end{array} \right.\\
\Rightarrow \left\{ \begin{array}{l}
y = 2\\
x = 3
\end{array} \right.\\
c)C = 2{x^2} + 2xy + 5{y^2} - 3x - 22y\\
= {x^2} + 2xy + {y^2} + {x^2} - 3x + \dfrac{9}{4}\\
+ 4{y^2} - 22y - \dfrac{9}{4}\\
= {\left( {x + y} \right)^2} + {\left( {x - \dfrac{3}{2}} \right)^2}\\
+ 4.\left( {{y^2} - \dfrac{{11}}{2}y + \dfrac{{121}}{{16}}} \right) - \dfrac{{65}}{2}\\
= {\left( {x + y} \right)^2} + {\left( {x - \dfrac{3}{2}} \right)^2} + 4{\left( {y - \dfrac{{11}}{4}} \right)^2} - \dfrac{{65}}{2}\\
\Rightarrow GTNN:C = \dfrac{{ - 65}}{2}
\end{array}$
