ĐK : \(x;y\ne0\)
Ta có : \(\left\{{}\begin{matrix}\frac{5}{x}+\frac{3}{y}=1\\\frac{2}{x}+\frac{1}{y}=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\frac{5}{x}+\frac{3}{y}=1\\\frac{6}{x}+\frac{3}{y}=-3\end{matrix}\right.\)
\(\Rightarrow\frac{1}{x}=-3-1=-4\Leftrightarrow x=-\frac{1}{4}\)
Thay vào , ta có : \(y=\frac{1}{7}\)
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