\(x.x=x\)
\(\Rightarrow x^2=x^1\)
\(\Rightarrow x^2-x^1=0\)
\(\Rightarrow x^1\left(x-1\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}x^1=0\\x-1=0\end{matrix}\right.\)
\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
\(x+y=x.y=\dfrac{x}{y}\)
Từ \(x.y=\dfrac{x}{y}\) ta có:
\(x=\dfrac{x}{y^2}\) \(\Rightarrow y^2=1\Rightarrow y=\pm1\)
Xét \(y=1\) ta có:
\(x+1=x=x\)
\(x=x+1\) (vô lí)
Xét \(y=-1\) ta có:
\(x-1=-x=-x\)
\(\Rightarrow x-1=-x\)
\(\Rightarrow2x=1\)
\(\Rightarrow x=\dfrac{1}{2}\)
Vậy \(y=-1\) và \(x=\dfrac{1}{2}\)
