1. Given $\sqrt{x^{2}-4}+\sqrt{2 x+y}=0$. Then the value of $x-y$ is ( ).
(A) 2
(B) 6
(C) 2 or -2
(D) 6 or -6
-、1.D.
Since $\sqrt{x^{2}-4} \geqslant 0, \sqrt{2 x+y} \geqslant 0$, therefore, it can only be that $\sqrt{x^{2}-4}=\sqrt{2 x+y}=0$.
Solving separately, we get $\left\{\begin{array}{l}x=2 \text { or }-2, \\ y=-2 x .\end{array}\right.$
Then $x-y=x-(-2 x)=3 x$, that is, $x-y$ is 6 or -6.