\(\left\{{}\begin{matrix}4xy+4\left(x^2+y^2\right)+\dfrac{3}{\left(x+y\right)^2}=7\\2x+\dfrac{1}{x+y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\left(x+y\right)^2+\left(x-y\right)^2+\dfrac{3}{\left(x+y\right)^2}=7\\\left(x+y\right)+\left(x-y\right)+\dfrac{1}{x+y}=1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}3\left[\left(x+y\right)+\dfrac{1}{x+y}\right]^2+\left(x-y\right)^2=13\\\left(x+y\right)+\left(x-y\right)+\dfrac{1}{x+y}=1\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y+\dfrac{1}{x+y}=a\\x-y=b\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}3a^2+b^2=13\\a+b=1\end{matrix}\right.\)
Đơn giản rồi nhé
