Từ \(\frac{x}{y}=x.y\Rightarrow x=x.y.y=x.y^2\Rightarrow y^2=\frac{x}{x}=1\Rightarrow y\in\left\{-1;1\right\}\)
+)y=-1
Ta có:3x+(-1)=x.(-1) (vì \(3x+y=x.y\))
=>3x-1=-x=>3x-(-x)=1=>4x=1=>x=\(\frac{1}{4}\)
+)y=1
Ta có:3x+1=x.1
=>3x+1=x=>3x-x=1=>2x=1=>x=\(\frac{1}{2}\))
Vậy \(x\in\left\{\frac{1}{2};\frac{1}{4}\right\};y\in\left\{1;-1\right\}\)