Đáp án:
\[{A_{\min }} = \frac{{35}}{4} \Leftrightarrow \left\{ \begin{array}{l}
x = - 2\\
y = \frac{1}{2}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = {x^2} + 4xy + 5{y^2} + 2x + 3y + 10\\
= \left( {{x^2} + 4xy + 4{y^2}} \right) + 2\left( {x + 2y} \right) + {y^2} - y + 10\\
= \left[ {{{\left( {x + 2y} \right)}^2} + 2\left( {x + 2y} \right) + 1} \right] + \left( {{y^2} - y + \frac{1}{4}} \right) + \frac{{35}}{4}\\
= {\left( {x + 2y + 1} \right)^2} + {\left( {y - \frac{1}{2}} \right)^2} + \frac{{35}}{4} \ge \frac{{35}}{4},\,\,\,\forall x,y\\
\Rightarrow {A_{\min }} = \frac{{35}}{4} \Leftrightarrow \left\{ \begin{array}{l}
x + 2y + 1 = 0\\
y - \frac{1}{2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = - 2\\
y = \frac{1}{2}
\end{array} \right.
\end{array}\)
