ĐKXĐ: ...
\(\left\{{}\begin{matrix}\dfrac{x+y}{xy}=-\dfrac{1}{2}\\\left(x+y\right)^2-2xy=5\end{matrix}\right.\)
Đặt \(\left\{{}\begin{matrix}x+y=u\\xy=v\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\dfrac{u}{v}=-\dfrac{1}{2}\\u^2-2v=5\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}v=-2u\\u^2-2v=5\end{matrix}\right.\) \(\Rightarrow u^2+4u=5\)
\(\Leftrightarrow...\)