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

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

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