Xin hay nhất (>_<)

Đáp án:

\[{A_{\max }} = 63 \Leftrightarrow \left\{ \begin{array}{l}
x =  - 7\\
y =  - 2
\end{array} \right.\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
A = 5 - 2{x^2} - 4{y^2} + 4xy - 8x - 12y\\
 = 5 - \left( {2{x^2} + 4{y^2} - 4xy + 8x + 12y} \right)\\
 = 5 - \left[ {\left( {{x^2} - 4xy + 4{y^2}} \right) + \left( { - 6x + 12y} \right) + \left( {{x^2} + 14x + 49} \right) - 49} \right]\\
 = 5 - \left[ {\left( {{x^2} - 2.x.2y + {{\left( {2y} \right)}^2}} \right) - 6.\left( {x - 2y} \right) + \left( {{x^2} + 2.x.7 + {7^2}} \right) - 49} \right]\\
 = 5 - \left[ {{{\left( {x - 2y} \right)}^2} - 6.\left( {x - 2y} \right) + {{\left( {x + 7} \right)}^2} - 49} \right]\\
 = 5 - \left\{ {\left[ {{{\left( {x - 2y} \right)}^2} - 6.\left( {x - 2y} \right) + 9} \right] + {{\left( {x + 7} \right)}^2} - 58} \right\}\\
 = 5 - \left\{ {\left[ {{{\left( {x - 2y} \right)}^2} - 2.\left( {x - 2y} \right).3 + {3^2}} \right] + {{\left( {x + 7} \right)}^2} - 58} \right\}\\
 = 5 - \left[ {{{\left( {x - 2y + 3} \right)}^2} + {{\left( {x + 7} \right)}^2} - 58} \right]\\
 = 5 - \left[ {{{\left( {x - 2y + 3} \right)}^2} + {{\left( {x + 7} \right)}^2}} \right] + 58\\
 = 63 - \left[ {{{\left( {x - 2y + 3} \right)}^2} + {{\left( {x + 7} \right)}^2}} \right]\\
{\left( {x - 2y + 3} \right)^2} \ge 0,\,\,\,\forall x,y\\
{\left( {x + 7} \right)^2} \ge 0,\,\,\,\forall x\\
 \Rightarrow {\left( {x - 2y + 3} \right)^2} + {\left( {x + 7} \right)^2} \ge 0,\,\,\,\forall x,y\\
 \Rightarrow 63 - \left[ {{{\left( {x - 2y + 3} \right)}^2} + {{\left( {x + 7} \right)}^2}} \right] \le 63,\,\,\,\forall x,y\\
 \Rightarrow A \le 63,\,\,\,\forall x,y\\
 \Rightarrow {A_{\max }} = 63 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - 2y + 3} \right)^2} = 0\\
{\left( {x + 7} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x =  - 7\\
y =  - 2
\end{array} \right.
\end{array}\)

Vậy \({A_{\max }} = 63 \Leftrightarrow \left\{ \begin{array}{l}
x =  - 7\\
y =  - 2
\end{array} \right.\)