$$\left\lbrace\begin{array}{l} 4\left(x^2+xy+y^2\right)+ \dfrac{3}{(x+y)^2}= \dfrac{85}{3}\\2x+ \dfrac{1}{x+y}= \dfrac{13}{3} \end{array}\right.$$

Đề bài:

\begin{equation} \label{eq:72.I} \left\lbrace\begin{array}{l} 4\left(x^2+xy+y^2\right)+ \dfrac{3}{(x+y)^2}= \dfrac{85}{3}\\2x+ \dfrac{1}{x+y}= \dfrac{13}{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}= \dfrac{85}{3} \\ x+y+ \dfrac{1}{x+y}+x-y= \dfrac{13}{3}\end{array} \right. \nonumber \\&\Leftrightarrow \left\lbrace \begin{array}{l} 3(x+y)^2+ \dfrac{3}{(x+y)^2}+6+(x-y)^2= \dfrac{103}{3} \\ x+y+ \dfrac{1}{x+y}+x-y= \dfrac{13}{3}\end{array} \right. \nonumber \\ \label{eq:72.1} &\Leftrightarrow \left\lbrace \begin{array}{l} 3\left(x+y+ \dfrac{1}{x+y}\right)^2+(x-y)^2= \dfrac{103}{3} \\ x+y+ \dfrac{1}{x+y}+x-y= \dfrac{13}{3}\end{array} \right. \end{align}

Đặt $x+y+ \dfrac{1}{x+y}=u$; $x-y=v$
\begin{align} \eqref{eq:72.1} &\Rightarrow \left\lbrace \begin{array}{l} 3u^2+v^2= \dfrac{103}{3} \\ u+v= \dfrac{13}{3}\end{array} \right. \nonumber \\ &\Leftrightarrow \left\lbrace \begin{array}{l} 3u^2+\left( \dfrac{13}{3}-u\right)^2= \dfrac{103}{3} \\ v= \dfrac{13}{3}-u\end{array} \right.\nonumber\\ &\Leftrightarrow \left\lbrace \begin{array}{l} 4u^2- \dfrac{26}{3}u- \dfrac{140}{9}=0 \\ v= \dfrac{13}{3}-u\end{array} \right.\nonumber \\ &\Leftrightarrow \left\lbrace \begin{array}{l} \left[\begin{array}{l} u= \dfrac{10}{3}\\u=- \dfrac{7}{6}\end{array} \right. \\ v= \dfrac{13}{3}-u \end{array} \right. \nonumber\\ &\Leftrightarrow \left[\begin{array}{l}\left\lbrace \begin{array}{l} u= \dfrac{10}{3}\\v=1 \end{array} \right. \\ \left\lbrace \begin{array}{l} u=- \dfrac{7}{6}\\v= \dfrac{11}{2} \end{array} \right. \end{array} \right. \nonumber \\ &\Rightarrow \left[\begin{array}{l}\left\lbrace \begin{array}{l} x+y+ \dfrac{1}{x+y}= \dfrac{10}{3}\\x-y=1 \end{array} \right. \\ \left\lbrace \begin{array}{l} x+y+ \dfrac{1}{x+y}=- \dfrac{7}{6}\\x-y= \dfrac{11}{2} \end{array} \right. \end{array} \right. \nonumber \\&\Leftrightarrow \left[ \begin{array}{l} \left\lbrace \begin{array}{l} (x+y)^2- \dfrac{10}{3}(x+y)+1=0 \\x-y=1 \end{array} \right. \\ \left\lbrace \begin{array}{l} (x+y)^2+ \dfrac{7}{6}(x+y)+1=0 \quad (\Delta=- \dfrac{95}{36}<0)\\x-y= \dfrac{11}{2} \end{array} \right. \end{array} \right. \nonumber \\&\Leftrightarrow \left\lbrace \begin{array}{l} \left[\begin{array}{l} x+y=3 \\ x+y= \dfrac{1}{3} \end{array}\right. \\x-y=1 \end{array} \right. \nonumber \\&\Leftrightarrow \left[ \begin{array}{l} \left\lbrace\begin{array}{l} x+y=3 \\ x-y=1 \end{array}\right. \\ \left\lbrace \begin{array}{l} x+y= \dfrac{1}{3}\\x-y=1 \end{array} \right. \end{array} \right. \nonumber \\&\Leftrightarrow \left[\begin{array}{l} \left\lbrace\begin{array}{l} x=2 \\ y=1 \end{array}\right. \\ \left\lbrace \begin{array}{l} x= \dfrac{2}{3}\\y=- \dfrac{1}{3} \end{array} \right. \end{array} \right. \nonumber \end{align}
$$S=\left\lbrace (2;1);\left( \dfrac{2}{3};- \dfrac{1}{3}\right) \right\rbrace$$