Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
1,\\
A = - {x^2} - 4{y^2} + 2xy + 2x + 10y - 3\\
= - \left( {{x^2} - 2xy + {y^2}} \right) + 2.\left( {x - y} \right) - \left( {3{y^2} - 12y + 12} \right) + 9\\
= - {\left( {x - y} \right)^2} + 2.\left( {x - y} \right) - 3.\left( {{y^2} - 4y + 4} \right) + 9\\
= - \left[ {{{\left( {x - y} \right)}^2} - 2.\left( {x - y} \right) + 1} \right] - 3.{\left( {y - 2} \right)^2} + 10\\
= - {\left( {x - y - 1} \right)^2} - 3.{\left( {y - 2} \right)^2} + 10\\
{\left( {x - y - 1} \right)^2} \ge 0,\,\,\,\,\forall x,y\\
{\left( {y - 2} \right)^2} \ge 0,\,\,\,\,\forall y\\
\Rightarrow 10 - {\left( {x - y - 1} \right)^2} - 3{\left( {y - 2} \right)^2} \le 10\\
\Rightarrow {A_{\max }} = 10 \Leftrightarrow \left\{ \begin{array}{l}
x - y - 1 = 0\\
y - 2 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = 2
\end{array} \right.\\
3,\\
C = 3{x^2} - 9x + 15\\
= 3.\left( {{x^2} - 3x + \frac{9}{4}} \right) + \frac{{33}}{4}\\
= 3.{\left( {x - \frac{3}{2}} \right)^2} + \frac{{33}}{4} \ge \frac{{33}}{4},\,\,\,\forall x\\
\Rightarrow {C_{\min }} = \frac{{33}}{4} \Leftrightarrow {\left( {x - \frac{3}{2}} \right)^2} = 0 \Leftrightarrow x = \frac{3}{2}\\
4,\\
D = 25 - 4{x^2} + 15 = 40 - 4{x^2} \le 40\\
\Rightarrow {D_{\max }} = 40 \Leftrightarrow {x^2} = 0 \Leftrightarrow x = 0\\
5,\\
E = {x^2} + 2{y^2} + 2xy - 2x + 5\\
= \left( {{x^2} + 2xy + {y^2}} \right) - \left( {2x + 2y} \right) + \left( {{y^2} + 2y + 1} \right) + 4\\
= {\left( {x + y} \right)^2} - 2.\left( {x + y} \right) + {\left( {y + 1} \right)^2} + 4\\
= \left[ {{{\left( {x + y} \right)}^2} - 2.\left( {x + y} \right) + 1} \right] + {\left( {y + 1} \right)^2} + 3\\
= {\left( {x + y - 1} \right)^2} + {\left( {y + 1} \right)^2} + 3 \ge 3,\,\,\,\forall x,y\\
\Rightarrow {E_{\min }} = 3 \Leftrightarrow \left\{ \begin{array}{l}
x + y - 1 = 0\\
y + 1 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 2\\
y = - 1
\end{array} \right.
\end{array}\)
