Đáp án:
$MinA = \dfrac{{289}}{{756}} \Leftrightarrow \left( {x;y} \right) = \left( {\dfrac{{23}}{{126}};\dfrac{{13}}{{63}}} \right)$
Giải thích các bước giải:
Ta có:
$\begin{array}{l}
A = 5{x^2} - 4xy + 5{y^2} - \dfrac{4}{3}y - x + \dfrac{{101}}{{180}}\\
5A = 25{x^2} - 20xy + 25{y^2} - \dfrac{{20}}{3}y - 5x + \dfrac{{101}}{{36}}\\
= {\left( {5x} \right)^2} - 2.5x\left( {2y + \dfrac{1}{2}} \right) + {\left( {2y + \dfrac{1}{2}} \right)^2} + 21{y^2} - \dfrac{{26}}{3}y + \dfrac{{101}}{{36}}\\
= {\left( {5x - 2y - \dfrac{1}{2}} \right)^2} + 21\left( {{y^2} - 2.y.\dfrac{{13}}{{63}} + \dfrac{{{{13}^2}}}{{{{63}^2}}}} \right) + \dfrac{{1445}}{{756}}\\
= {\left( {5x - 2y - \dfrac{1}{2}} \right)^2} + 21{\left( {y - \dfrac{{13}}{{63}}} \right)^2} + \dfrac{{1445}}{{756}}
\end{array}$
Lại có:
$\begin{array}{l}
{\left( {5x - 2y - \dfrac{1}{2}} \right)^2} \ge 0;{\left( {y - \dfrac{{13}}{{63}}} \right)^2} \ge 0,\forall x,y\\
\Rightarrow 5A \ge \dfrac{{1445}}{{756}}\\
\Rightarrow A \ge \dfrac{{289}}{{756}}\\
\Rightarrow MinA = \dfrac{{289}}{{756}}
\end{array}$
Dấu bằng xảy ra:
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
5x - 2y - \dfrac{1}{2} = 0\\
y - \dfrac{{13}}{{63}} = 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
x = \dfrac{{23}}{{126}}\\
y = \dfrac{{13}}{{63}}
\end{array} \right.
\end{array}$
Vậy $MinA = \dfrac{{289}}{{756}} \Leftrightarrow \left( {x;y} \right) = \left( {\dfrac{{23}}{{126}};\dfrac{{13}}{{63}}} \right)$
