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Ta có:
$ \begin{array}{l}
5{x^2} + 8xy + 5{y^2} = 36
\Leftrightarrow \left( {{x^2} + {y^2}} \right) + 4\left( {x + y} \right) = 36
\Leftrightarrow S = 36 - 4{\left( {x + y} \right)^2}\\
\Rightarrow S \le 36
\end{array} $
Dấu “=” xảy ra
$ \Leftrightarrow \left\{ \begin{array}{l}
x + y = 0\\
5{x^2} + 8xy + 5{y^2} = 36
\end{array} \right. $ $ \begin{array}{l}
\Leftrightarrow x = 3\sqrt 2 ,y = - 3\sqrt 2 \,\,\,V\,\,\,x = - 3\sqrt 2 ,y = 3\sqrt 2 .
\Rightarrow \max \,S = 36
\end{array} $
Mặt khác, ta có thể viết:
$ \begin{array}{l}
5{x^2} + 8xy + {y^2} = 36
\Leftrightarrow 9\left( {{x^2} + {y^2}} \right) - 4{\left( {x - y} \right)^2} = 36
\Leftrightarrow 9S = 36 + 4{\left( {x - y} \right)^2}\\
\Rightarrow 9S \ge 36\\
\Rightarrow S \ge 4
\end{array} $
Dấu “=” xảy ra $ \Leftrightarrow \left\{ \begin{array}{l}
x - y = 0\\
5{x^2} + 8xy + 5{y^2} = 36
\end{array} \right. $ $ \begin{array}{l}
\Leftrightarrow x = y = \sqrt 2 \
\Rightarrow \min \,\,S = \sqrt 4
\end{array} $
Vậy ta có: $ \left[ \begin{array}{l}
\max\,S = 36\\
\min \,S = 4
\end{array} \right. $
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