a,\(x+y=xy\)
\(\)\(\Rightarrow x+y-xy=0\)
\(\Rightarrow x+y-xy-1=-1\)
\(\Rightarrow\left(x-xy\right)+\left(y-1\right)=-1\)
\(\Rightarrow x\left(1-y\right)-\left(1-y\right)=-1\)
\(\Rightarrow\left(x-1\right)\left(1-y\right)=-1\)
\(\Rightarrow x-1;1-y\inƯ\left(-1\right)=\left\{1;-1\right\}\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=1\\x-1=-1\\1-y=1\\1-y=-1\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=2\\x=0\\y=0\\y=2\end{matrix}\right.\)
Vậy có 4 trường hợp:
TH1:\(x=2;y=0\)
TH2:\(x=0;y=2\)
TH3:\(x=0;y=0\)
TH4:\(x=2;y=2\)
a)\(x+y=xy\)
\(\Leftrightarrow x+y-xy=0\)
\(\Leftrightarrow x+y-xy-1=-1\)
\(\Leftrightarrow x-xy+y-1=-1\)
\(\Leftrightarrow x\left(1-y\right)+\left(y-1\right)=-1\)
\(\Leftrightarrow x\left(1-y\right)-\left(1-y\right)=-1\)
\(\Leftrightarrow\left(1-y\right)\left(x-1\right)=-1\)
\(\Rightarrow\left\{{}\begin{matrix}1-y\\x-1\end{matrix}\right.\inƯ\left(-1\right)=\left\{-1;1\right\}\)
+\(\left\{{}\begin{matrix}1-y=-1\\x-1=-1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=2\\x=0\end{matrix}\right.\)
+\(\left\{{}\begin{matrix}1-y=1\\x-1=1\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}y=0\\x=2\end{matrix}\right.\)
Vậy..............
