1. $x, y$ are real numbers, and $\left(x+\sqrt{x^{2}+1}\right)(y+$ $\left.\sqrt{y^{2}+1}\right)=1$. Then $x+y=$ $\qquad$ .
二、1.0.
Multiply both sides of the original equation by $\sqrt{x^{2}+1}-x$, we get $y+\sqrt{y^{2}+1}=\sqrt{x^{2}+1}-x$; multiply both sides of the original equation by $\sqrt{y^{2}+1}-y$, we get $x+\sqrt{x^{2}+1}=\sqrt{y^{2}+1}-y$. Adding the two equations immediately yields $x+y=0$.