\(x+\sqrt{x^2+1}=y+\sqrt{y^2-1}\) (1)
\(\Rightarrow\dfrac{1}{\sqrt{x^2+1}-x}=\dfrac{1}{y-\sqrt{y^2-1}}\)
\(\Rightarrow\sqrt{x^2+1}-x=y-\sqrt{y^2-1}\) (2)
Cộng vế (1) và (2):
\(2\sqrt{x^2+1}=2y\Rightarrow y=\sqrt{x^2+1}\)
Thế vào pt \(x^2+y^2-xy=1\)
\(\Rightarrow x^2+x^2+1-x\sqrt{x^2+1}=1\)
\(\Leftrightarrow2x^2-x\sqrt{x^2+1}=0\)
\(\Rightarrow\left[{}\begin{matrix}x=0\Rightarrow y=1\\2x=\sqrt{x^2+1}\left(3\right)\end{matrix}\right.\)
Xét (3) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\4x^2=x^2+1\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2=\dfrac{1}{3}\end{matrix}\right.\)
\(\Rightarrow x=\dfrac{1}{\sqrt{3}}\Rightarrow y=\dfrac{2}{\sqrt{3}}\)
