ĐKXĐ: ...
\(\Leftrightarrow\left\{{}\begin{matrix}3\left(x+y\right)^2+\frac{3}{\left(x+y\right)^2}+\left(x-y\right)^2=7\\x+y+\frac{1}{x+y}+x-y=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\left(x+y+\frac{1}{x+y}\right)^2+\left(x-y\right)^2=13\\x+y+\frac{1}{x+y}+x-y=3\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y+\frac{1}{x+y}=a\\x-y=b\end{matrix}\right.\) với \(\left|a\right|\ge2\)
\(\Rightarrow\left\{{}\begin{matrix}3a^2+b^2=13\\a+b=3\end{matrix}\right.\) \(\Rightarrow3a^2+\left(3-a\right)^2=13\)
\(\Leftrightarrow2a^2-3a-2=0\Rightarrow\left[{}\begin{matrix}a=2\Rightarrow b=1\\a=-\frac{1}{2}\left(l\right)\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x+y+\frac{1}{x+y}=2\\x-y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y\right)^2-2\left(x+y\right)+1=0\\x-y=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\left(x+y-1\right)^2=0\\x-y=1\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x+y=1\\x-y=1\end{matrix}\right.\)
\(\Leftrightarrow...\)