Đáp án:
\[{A_{\min }} = 2 \Leftrightarrow \left\{ \begin{array}{l}
x = 4\\
y = \frac{{16}}{3}
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = 5{x^2} + 9{y^2} - 12xy + 24x - 48y + 82\\
= \left( {4{x^2} - 12xy + 9{y^2}} \right) + 16\left( {2x - 3y} \right) + \left( {{x^2} - 8x + 16} \right) + 66\\
= {\left( {2x - 3y} \right)^2} + 16\left( {2x - 3y} \right) + {\left( {x - 4} \right)^2} + 66\\
= \left[ {{{\left( {2x - 3y} \right)}^2} + 16\left( {2x - 3y} \right) + 64} \right] + {\left( {x - 4} \right)^2} + 2\\
= {\left( {2x - 3y + 8} \right)^2} + {\left( {x - 4} \right)^2} + 2 \ge 2,\,\,\,\,\forall x,y\\
A = 2 \Leftrightarrow \left\{ \begin{array}{l}
2x - 3y + 8 = 0\\
x - 4 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 4\\
y = \frac{{16}}{3}
\end{array} \right.
\end{array}\)
Vậy \({A_{\min }} = 2 \Leftrightarrow \left\{ \begin{array}{l}
x = 4\\
y = \frac{{16}}{3}
\end{array} \right.\)
