Đề bài: (Một bài dễ - nhưng đánh giá cao ý tưởng)
Giải hệ phương trình: \ \left\{ \begin{array}{l}
\frac{{{x^2}}}{{{{\left( {y + 1} \right)}^2}}} + \frac{{{y^2}}}{{{{\left( {x + 1} \right)}^2}}} = \frac{1}{2}\\
3xy = x + y + 1
\end{array} \right.\left( * \right).
Giải:
Ta có: \ \left( * \right) \Leftrightarrow \left\{ \begin{array}{l}
\frac{{{x^2}}}{{{{\left( {y + 1} \right)}^2}}} + \frac{{{y^2}}}{{{{\left( {x + 1} \right)}^2}}} = \frac{1}{2}\\
4xy = xy + x + y + 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\frac{{{x^2}}}{{{{\left( {y + 1} \right)}^2}}} + \frac{{{y^2}}}{{{{\left( {x + 1} \right)}^2}}} = \frac{1}{2}\\
4xy = \left( {x + 1} \right)\left( {y + 1} \right)
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\frac{{{x^2}}}{{{{\left( {y + 1} \right)}^2}}} + \frac{{{y^2}}}{{{{\left( {x + 1} \right)}^2}}} = \frac{1}{2}\\
\frac{x}{{y + 1}}.\frac{y}{{x + 1}} = \frac{1}{4}
\end{array} \right.
Áp dụng BĐT Cauchy cho 2 số \ {\left( {\frac{x}{{y + 1}}} \right)^2}\& {\left( {\frac{y}{{x + 1}}} \right)^2} ta có:
\ \frac{1}{2} = {\left( {\frac{x}{{y + 1}}} \right)^2} + {\left( {\frac{y}{{x + 1}}} \right)^2} \ge 2\left| {\frac{x}{{y + 1}}.\frac{y}{{x + 1}}} \right| = \frac{1}{2}.
Dấu "=" xảy ra \ \Leftrightarrow \left[ \begin{array}{l}
\frac{x}{{y + 1}} = \frac{y}{{x + 1}} = \frac{1}{2} \Leftrightarrow \frac{{x + y}}{{x + y + 2}} = \frac{1}{2} \Leftrightarrow x + y = 2 \Rightarrow \left\{ \begin{array}{l}
x + y = 2\\
xy = 1
\end{array} \right. \Leftrightarrow \left( {x;y} \right) = \left( {1;1} \right)\\
\frac{x}{{y + 1}} = \frac{y}{{x + 1}} = - \frac{1}{2} \Leftrightarrow \frac{{x + y}}{{x + y + 2}} = - \frac{1}{2} \Rightarrow \left\{ \begin{array}{l}
x + y = - \frac{2}{3}\\
xy = \frac{1}{9}
\end{array} \right. \Leftrightarrow \left( {x;y} \right) = \left( { - \frac{1}{3}; - \frac{1}{3}} \right)
\end{array} \right.
Vậy \ S = \left\{ {\left( {1;1} \right),\left( { - \frac{1}{3}; - \frac{1}{3}} \right)} \right\}.
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