ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}8x^2+8y^2+4xy-13+\frac{5}{\left(x+y\right)^2}=0\\2x+\frac{1}{x+y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5\left(x+y\right)^2+\frac{5}{\left(x+y\right)^2}+10+3\left(x-y\right)^2=23\\x+y+\frac{1}{x+y}+x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}5\left(x+y+\frac{1}{x+y}\right)^2+\left(x-y\right)^2=23\\x+y+\frac{1}{x+y}+x-y=1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y+\frac{1}{x+y}=a\\x-y=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}5a^2+b^2=23\\a+b=1\end{matrix}\right.\) \(\Rightarrow5a^2+\left(1-a\right)^2-23=0\)