Question

Tìm x€Qvà y nhé

x.x=x

x+y=x.y=x:y(x=0,5;y=-1)

Answer 1

\(x.x=x\)

\(\Rightarrow x^2=x^1\)

\(\Rightarrow x^2-x^1=0\)

\(\Rightarrow x^1\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x^1=0\\x-1=0\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

\(x+y=x.y=\dfrac{x}{y}\)

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 \(x=\dfrac{1}{2}\)