1. Given $\sqrt{2 \sqrt{3}-3}=\sqrt{\sqrt{3} x}-\sqrt{\sqrt{3} y}(x, y$ are rational numbers). Then $x-y=(\quad)$.
(A) $\frac{3}{2}$
(B) $\frac{1}{2}$
(C) 1
(D) 2
- 1. C.

From the given, we have
$$
\sqrt{x}-\sqrt{y}=\sqrt{2-\sqrt{3}},
$$

which means $\sqrt{x}-\sqrt{y}=\sqrt{\frac{4-2 \sqrt{3}}{2}}=\frac{\sqrt{3}-1}{\sqrt{2}}=\sqrt{\frac{3}{2}}-\sqrt{\frac{1}{2}}$.
Squaring both sides, we get
$$
x+y-2 \sqrt{x y}=\frac{3}{2}+\frac{1}{2}-2 \sqrt{\frac{3}{2} \times \frac{1}{2}}.
$$

Since $x$ and $y$ are both rational numbers, we have $x y=\frac{3}{2} \times \frac{1}{2}, x+y=\frac{3}{2}+\frac{1}{2}$.
Given that $x>y$, then $x=\frac{3}{2}, y=\frac{1}{2}$.
Therefore, $x-y=1$.