Giải:
Ta có: \(x-y=xy\)
\(\Rightarrow x=xy+y\)
\(\Rightarrow x=\left(x+1\right)y\)
Mà \(x-y=x:y\)
\(\Rightarrow x+1=x:y\)
Ta có: \(x:y=x-y\)
\(\Rightarrow x+1=x-y\)
\(\Rightarrow y=-1\)
\(x-y=xy\)
\(\Rightarrow x-\left(-1\right)=x\left(-1\right)\)
\(\Rightarrow x+1=-x\)
\(\Rightarrow x+x=-1\)
\(\Rightarrow2x=-1\)
\(\Rightarrow x=\frac{-1}{2}\)
Vậy cặp số \(\left(x;y\right)\) là \(\left(-1;\frac{-1}{2}\right)\)
\(\Rightarrow y=-1\)
Ta có:
x - y = xy => x = xy + y = y.(x + 1) (1)
=> x : y = x + 1 = x - y (theo đề bài)
=> y = -1
Thay y = -1 vào (1) ta có: x = -1(x + 1) = -x - 1
=> x + x = -1 = 2x
=> \(x=\frac{-1}{2}\)
Vậy \(x=\frac{-1}{2};y=-1\)
