ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{2x}+\dfrac{1}{y}=\dfrac{3}{2}\\\dfrac{1}{x}+\dfrac{2}{y}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\dfrac{1}{x}+\dfrac{2}{y}=3\\\dfrac{1}{x}+\dfrac{2}{y}=3\left(luôn.đúng\right)\end{matrix}\right.\)
Vậy \(\left\{{}\begin{matrix}x,y\in R\\x,y\ne0\end{matrix}\right.\)