Đáp án:
\[{A_{\min }} = - 17 \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = 4
\end{array} \right.\]
Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
A = {x^2} + 2{y^2} - 2xy + 2x - 10y\\
= \left( {{x^2} - 2xy + {y^2}} \right) + \left( {2x - 2y} \right) + \left( {{y^2} - 8y + 16} \right) - 16\\
= {\left( {x - y} \right)^2} + 2.\left( {x - y} \right) + {\left( {y - 4} \right)^2} - 16\\
= \left[ {{{\left( {x - y} \right)}^2} + 2.\left( {x - y} \right) + 1} \right] + {\left( {y - 4} \right)^2} - 17\\
= {\left[ {\left( {x - y} \right) + 1} \right]^2} + {\left( {y - 4} \right)^2} - 17\\
= {\left( {x - y + 1} \right)^2} + {\left( {y - 4} \right)^2} - 17\\
{\left( {x - y + 1} \right)^2} \ge 0,\,\,\,\forall x,y\\
{\left( {y - 4} \right)^2} \ge 0,\,\,\,\forall y\\
\Rightarrow {\left( {x - y + 1} \right)^2} + {\left( {y - 4} \right)^2} \ge 0,\,\,\,\forall x,y\\
\Rightarrow A = {\left( {x - y + 1} \right)^2} + {\left( {y - 4} \right)^2} - 17 \ge - 17,\,\,\,\forall x,y\\
\Rightarrow {A_{\min }} = - 17 \Leftrightarrow \left\{ \begin{array}{l}
{\left( {x - y + 1} \right)^2} = 0\\
{\left( {y - 4} \right)^2} = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x - y + 1 = 0\\
y - 4 = 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = 4
\end{array} \right.
\end{array}\)
Vậy \({A_{\min }} = - 17 \Leftrightarrow \left\{ \begin{array}{l}
x = 3\\
y = 4
\end{array} \right.\)
