Vì \(x.y=x-y\)
\(\Rightarrow x.y+2x+y=x-y+2x+y\)
\(\Rightarrow\) \(x-y+2x+y=1\)
\(\Leftrightarrow\left(x+2x\right)+\left(-y+y\right)=1\)
\(\Leftrightarrow3x+0=1\)
\(\Rightarrow3x=1\)
\(\Rightarrow x=\dfrac{1}{3}\)
Thay \(x=\dfrac{1}{3}\) ta có:
\(\dfrac{1}{3}.y+2.\dfrac{1}{3}+y=1\)
\(\Leftrightarrow\left(\dfrac{1}{3}.y+y\right)+2.\dfrac{1}{3}=1\)
\(\Leftrightarrow y\left(\dfrac{1}{3}+1\right)+\dfrac{2}{3}=1\)
\(\Leftrightarrow y.\dfrac{4}{3}+\dfrac{2}{3}=1\)
\(\Rightarrow\dfrac{4y}{3}=1-\dfrac{2}{3}\)
\(\Rightarrow\dfrac{4y}{3}=\dfrac{1}{3}\)
\(\Rightarrow4y=1\)
\(\Rightarrow y=\dfrac{1}{4}\)
Vậy \(x=\dfrac{1}{3}\) và \(y=\dfrac{1}{4}\)
