\(\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{3}\)
\(\Leftrightarrow3y-3x-xy=0\) \(\Leftrightarrow y\left(3-x\right)+3\left(3-x\right)=9\) \(\Leftrightarrow\left(y+3\right)\left(3-x\right)=9\)
Do \(y\in Z^+\Rightarrow y+3\inƯ\left(9\right),y+3>3\) \(\Rightarrow y+3=9\) \(\Rightarrow3-x=1\)
\(\Rightarrow\left\{{}\begin{matrix}x=2\\y=6\end{matrix}\right.\)
ta có \(\dfrac{1}{x}-\dfrac{1}{y}=\dfrac{1}{3}\left(x,y\ne0\right)< =>\dfrac{y-x}{xy}=\dfrac{1}{3}\)
=>\(3\left(y-x\right)=xy< =>3y-3x-xy=0\)
<=>\(3y-xy+9-3x-9=0\)
<=>\(y\left(3-x\right)+3\left(3-x\right)=9< =>\left(y+3\right)\left(3-x\right)=9\)
=>(y+3)(3-x)\(\inƯ\left(9\right)=\left(1,-1,3,-3,9,-9\right)\)
vì x,y nguyên dương nên (y+3)(3-x)=(1,3,9)
* với \(\left\{{}\begin{matrix}y+3=1\\3-x=9\end{matrix}\right.< =>\left\{{}\begin{matrix}y=-2\\x=-6\end{matrix}\right.\)(loại)
*với\(\left\{{}\begin{matrix}y+3=9\\3-x=1\end{matrix}\right.< =>\left\{{}\begin{matrix}y=6\\x=2\end{matrix}\right.\)(thỏa mãn)
*với\(\left\{{}\begin{matrix}y+3=3\\3-x=3\end{matrix}\right.< =>\left\{{}\begin{matrix}y=0\\x=0\end{matrix}\right.\)(loại vì x,y\(\ne\)0)
vậy (x,y)=(2,6)
