\(x+y-xy+1=0\)
\(x+y-xy-1=-2\)
\(\Rightarrow x\left(1-y\right)-1\left(1-y\right)=-2\)
\(\Rightarrow\left(x-1\right)\left(1-y\right)=-2\)
\(\Rightarrow x-1;1-y\in U\left(-2\right)\)
\(U\left(-2\right)=\left\{\pm1;\pm2\right\}\)
\(\Rightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x-1=1\Rightarrow x=2\\1-y=-2\Rightarrow y=3\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-1\Rightarrow x=0\\1-y=2\Rightarrow y=-1\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=2\Rightarrow x=3\\1-y=-1\Rightarrow y=2\end{matrix}\right.\\\left\{{}\begin{matrix}x-1=-2\Rightarrow x=-1\\1-y=1\Rightarrow y=0\end{matrix}\right.\end{matrix}\right.\)\
\(\dfrac{2}{x}-\dfrac{1}{9}=\dfrac{y}{3}\)
\(\Rightarrow\dfrac{2}{x}-\dfrac{1}{9}=\dfrac{3y}{9}\)
\(\Rightarrow\dfrac{2}{x}=\dfrac{3y}{9}+\dfrac{1}{9}\)
\(\Rightarrow\dfrac{2}{x}=\dfrac{3y+1}{9}\)
\(\Rightarrow x\left(3y+1\right)=18\)
\(\Rightarrow x;3y+1\in U\left(18\right)\)
Xét ước như bài trên
\(3x+3y-xy=0\)
\(\Rightarrow3x+3y-xy-9=-9\)
\(\Rightarrow x\left(3-y\right)-3\left(3-y\right)=-9\)
\(\Rightarrow\left(x-3\right)\left(3-y\right)=-9\)
\(\Rightarrow x-3;3-y\in U\left(9\right)\)
Xét ước ~~~
