$$\left\lbrace\begin{array}{l} 4xy+4\left(x^2+y^2\right)+ \dfrac{3}{(x+y)^2}=7\\2x+ \dfrac{1}{x+y}=3 \end{array}\right. $$

Đề bài:

\begin{equation} \label{eq:73.I}\left\lbrace\begin{array}{l} 4xy+4\left(x^2+y^2\right)+ \dfrac{3}{(x+y)^2}=7\\2x+ \dfrac{1}{x+y}=3 \end{array}\right. \end{equation}

Lời giải:

\begin{align} \eqref{eq:72.I} &\Leftrightarrow \left\lbrace \begin{array}{l} 3(x^2+2xy+y^2)+(x^2-2xy+y^2)+ \dfrac{3}{(x+y)^2}=7 \\ x+y+ \dfrac{1}{x+y}+x-y=3\end{array} \right. \nonumber \\ &\Leftrightarrow \left\lbrace \begin{array}{l} 3(x+y)^2+ \dfrac{3}{(x+y)^2}+6+(x-y)^2=13 \\ x+y+ \dfrac{1}{x+y}+x-y=3\end{array} \right. \nonumber \\ \label{eq:73.1}&\Leftrightarrow \left\lbrace \begin{array}{l} 3\left(x+y+ \dfrac{1}{x+y}\right)^2+(x-y)^2=13 \\ x+y+ \dfrac{1}{x+y}+x-y=3\end{array} \right. \end{align}

Đặt $x+y+\dfrac{1}{x+y}=u$; $x-y=v$
\begin{align} \eqref{eq:73.1} &\Rightarrow \left\lbrace \begin{array}{l} 3u^2+v^2=13 \\ u+v=3\end{array} \right. \nonumber \\ &\Leftrightarrow \left\lbrace \begin{array}{l} 3u^2+\left(3-u\right)^2=13 \\ v=3-u\end{array} \right.\nonumber \\ &\Leftrightarrow \left\lbrace \begin{array}{l} 2u^2-3u-2=0 \quad \left( \Delta_u=25 >0 \right) \\ v=3-u\end{array} \right.\nonumber\\ &\Leftrightarrow \left[\begin{array}{l}\left\lbrace \begin{array}{l} u=2\\v=1 \end{array} \right. \\ \left\lbrace \begin{array}{l} u=- \dfrac{1}{2}\\v= \dfrac{7}{2} \end{array} \right. \end{array} \right. \nonumber \\ &\Rightarrow \left[\begin{array}{l}\left\lbrace \begin{array}{l} x+y+ \dfrac{1}{x+y}=2\\x-y=1 \end{array} \right. \\ \left\lbrace \begin{array}{l} x+y+ \dfrac{1}{x+y}=- \dfrac{1}{2}\\x-y= \dfrac{7}{2} \end{array} \right. \end{array} \right. \nonumber \\ &\Leftrightarrow \left[ \begin{array}{l} \left\lbrace \begin{array}{l} (x+y)^2-2(x+y)+1=0 \\x-y=1 \end{array} \right. \\ \left\lbrace \begin{array}{l} (x+y)^2+ \dfrac{1}{2}(x+y)+1=0 \quad (\Delta=- \dfrac{15}{4}<0)\\x-y= \dfrac{7}{2} \end{array} \right. \end{array} \right. \nonumber \\ &\Leftrightarrow \left\lbrace \begin{array}{l} x+y=1 \\x-y=1 \end{array} \right. \nonumber \\ &\Leftrightarrow \left\lbrace\begin{array}{l} x=1 \\ y=0 \end{array} \right. \nonumber \end{align}
$$S=\left\lbrace (1;0) \right\rbrace$$