Tìm GTNN của: A=5x^2-4xy+5y^2-4/3y-x+101/180

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 $MinA = \dfrac{{289}}{{756}} \Leftrightarrow \left( {x;y} \right) = \left( {\dfrac{{23}}{{126}};\dfrac{{13}}{{63}}} \right)$