Giải thích các bước giải:

Ta có:

\(\begin{array}{l}
a,\\
A =  - {x^2} - 2{y^2} + 12x - 4y + 7\\
 = \left( { - {x^2} + 12x - 36} \right) + \left( { - 2{y^2} - 4y - 2} \right) + 45\\
 =  - \left( {{x^2} - 12x + 36} \right) - 2.\left( {{y^2} + 2y + 1} \right) + 45\\
 = 45 - \left( {{x^2} - 2.x.6 + {6^2}} \right) - 2.\left( {{y^2} + 2.y.1 + 1} \right)\\
 = 45 - {\left( {x - 6} \right)^2} - 2.{\left( {y + 1} \right)^2} \le 45,\,\,\,\forall x,y\\
 \Rightarrow {A_{\max }} = 45 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - 6} \right)^2} = 0\\
{\left( {y + 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 6\\
y =  - 1
\end{array} \right.\\
b,\\
B =  - {x^2} - 5{y^2} - 4xy + 2x - 2y - 5\\
 = \left( { - {x^2} - 4xy - 4{y^2}} \right) + \left( {2x + 4y} \right) + \left( { - {y^2} - 6y - 9} \right) + 4\\
 =  - \left( {{x^2} + 4xy + 4{y^2}} \right) + 2.\left( {x + 2y} \right) - \left( {{y^2} + 6y + 9} \right) + 4\\
 =  - {\left( {x + 2y} \right)^2} + 2.\left( {x + 2y} \right) - 1 - {\left( {y + 3} \right)^2} + 5\\
 =  - \left[ {{{\left( {x + 2y} \right)}^2} - 2.\left( {x + 2y} \right) + 1} \right] - {\left( {y + 3} \right)^2} + 5\\
 = 5 - {\left( {x + 2y - 1} \right)^2} - {\left( {y + 3} \right)^2} \le 5,\,\,\,\forall x,y\\
 \Rightarrow {B_{\max }} = 5 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x + 2y - 1} \right)^2} = 0\\
{\left( {y + 3} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 7\\
y =  - 3
\end{array} \right.\\
a,\\
A = {x^2} + 2{y^2} - 2xy + 4x - 2y + 12\\
 = \left( {{x^2} - 2xy + {y^2}} \right) + \left( {4x - 4y} \right) + \left( {{y^2} + 2y + 1} \right) + 11\\
 = {\left( {x - y} \right)^2} + 4.\left( {x - y} \right) + 4 + {\left( {y + 1} \right)^2} + 7\\
 = \left[ {{{\left( {x - y} \right)}^2} + 4.\left( {x - y} \right) + 4} \right] + {\left( {y + 1} \right)^2} + 7\\
 = {\left( {x - y + 2} \right)^2} + {\left( {y + 1} \right)^2} + 7 \ge 7,\,\,\,\forall x,y\\
 \Rightarrow {A_{\min }} = 7 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - y + 2} \right)^2} = 0\\
{\left( {y + 1} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x =  - 3\\
y =  - 1
\end{array} \right.
\end{array}\)