2.9. (England, 70). Solve the equation

$$
\sqrt{2 \sqrt{3}-3}=\sqrt{x \sqrt{3}}-\sqrt{y \sqrt{3}}
$$

in rational numbers.
2.9. Let the numbers $x, y \in Q$ satisfy the equation. Then the following equalities hold

$$
\begin{gathered}
2 \sqrt{3}-3=x \sqrt{3}+y \sqrt{3}-2 \sqrt{3 x y} \\
(x+y-2) \sqrt{ } \overline{3}=2 \sqrt{3 x y}-3
\end{gathered}
$$

Since

$$
(x+y-2)^{2} \cdot 3=9+12 x y-12 \sqrt{3 x y}
$$

the number $\sqrt{3 x y}$ is rational, and therefore the equalities

$$
\begin{aligned}
x+y-2 & =0 \\
2 \sqrt{3 x y}-3 & =0
\end{aligned}
$$

hold (otherwise the number $\sqrt{3}=(2 \sqrt{3 x y}-3) /(x+y-2)$ would be rational). Therefore, the numbers $x, y$ satisfy the equalities $x+y=2$, $x y=3 / 4$, i.e., they are the roots of the equation

$$
t^{2}-2 t+3 / 4=0
$$

Since $x>y$, the original equation can have only one solution $x=3 / 2, y=1 / 2$, which, as a check shows, indeed satisfies the equation.