\(\frac{x+1}{2}-\frac{3}{5}=\frac{1}{2y}\Leftrightarrow\frac{5x-1}{10}=\frac{1}{2y}\)
\(\Leftrightarrow2y\left(5x-1\right)=10\Leftrightarrow y\left(5x-1\right)=5\)
TH1: \(\left\{{}\begin{matrix}y=5\\5x-1=1\end{matrix}\right.\) \(\Rightarrow x=\frac{2}{5}\left(l\right)\)
TH2: \(\left\{{}\begin{matrix}y=1\\5x-1=5\end{matrix}\right.\) \(\Rightarrow x=\frac{6}{5}\left(l\right)\)
TH3: \(\left\{{}\begin{matrix}y=-1\\5x-1=-5\end{matrix}\right.\) \(\Rightarrow x=-\frac{4}{5}\left(l\right)\)
TH4: \(\left\{{}\begin{matrix}y=-5\\5x-1=-1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=0\\y=-5\end{matrix}\right.\)
Vậy \(\left(x;y\right)=\left(0;-5\right)\)
