Bổ sung: \(x,y\in Z;x;y\ne0\)
\(\dfrac{2}{x}+\dfrac{3}{y}=1\)
\(\Leftrightarrow\dfrac{2}{x}=1-\dfrac{3}{y}\)
\(\Leftrightarrow\dfrac{2}{x}=\dfrac{y-3}{y}\)
\(\Leftrightarrow\dfrac{x}{2}=\dfrac{y}{y-3}\)
\(\Leftrightarrow x=\dfrac{2y}{y-3}\)
\(\Leftrightarrow x=2+\dfrac{6}{y-3}\)
Để x;y nguyên thì \(y-3\in U\left(6\right)=\left\{\pm1;\pm3;\pm6\right\}\)
TH1: \(y-3=1\Rightarrow y=4\)
\(x=8\)
Xét tất cả ta được các nghiệm
\(\left\{{}\begin{matrix}x=8\\y=4\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=-4\\y=2\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=3\\y=6\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=0\\y=0\end{matrix}\right.\) ( ktm ) \(\left\{{}\begin{matrix}x=3\\y=9\end{matrix}\right.\) \(\left\{{}\begin{matrix}x=1\\y=-3\end{matrix}\right.\)
