1: ĐKXĐ: \(\left\{{}\begin{matrix}x\ne0\\y\ne0\end{matrix}\right.\)
\(\left\{{}\begin{matrix}\dfrac{1}{2x}-\dfrac{2}{y}=3\\\dfrac{4}{x}+\dfrac{5}{6y}=\dfrac{1}{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{2}\cdot\dfrac{1}{x}-\dfrac{1}{y}\cdot2=3\\4\cdot\dfrac{1}{x}+\dfrac{5}{6}\cdot\dfrac{1}{y}=\dfrac{1}{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}4\cdot\dfrac{1}{x}-16\cdot\dfrac{1}{y}=24\\4\cdot\dfrac{1}{x}+\dfrac{5}{6}\cdot\dfrac{1}{y}=\dfrac{1}{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{y}\left(-16-\dfrac{5}{6}\right)=24-\dfrac{1}{6}\\\dfrac{4}{x}+\dfrac{5}{6y}=\dfrac{1}{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}\dfrac{1}{y}=-\dfrac{143}{101}\\\dfrac{4}{x}+\dfrac{5}{6y}=\dfrac{1}{6}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{101}{143}\\\dfrac{4}{x}=\dfrac{1}{6}-\dfrac{5}{6}\cdot\dfrac{1}{y}=\dfrac{136}{101}\end{matrix}\right.\)
=>\(\left\{{}\begin{matrix}y=-\dfrac{101}{143}\left(nhận\right)\\x=\dfrac{101\cdot4}{136}=\dfrac{101}{34}\left(nhận\right)\end{matrix}\right.\)