2.301. $\frac{\sqrt{5-2 \sqrt{6}} \cdot(5+2 \sqrt{6})(49-20 \sqrt{6})}{\sqrt{27}-3 \sqrt{18}+3 \sqrt{12}-\sqrt{8}}=1$.
Solution.

$$
\begin{aligned}
& \frac{\sqrt{5-2 \sqrt{6}} \cdot(5+2 \sqrt{6})(49-20 \sqrt{6})}{\sqrt{27}-3 \sqrt{18}+3 \sqrt{12}-\sqrt{8}}= \\
& =\frac{\sqrt{3-2 \sqrt{3 \cdot 2}+2} \cdot(3+2 \sqrt{3 \cdot 2}+2)(49-20 \sqrt{6})}{\sqrt{9 \cdot 3}-3 \sqrt{9 \cdot 2}+3 \sqrt{4 \cdot 3}-\sqrt{4 \cdot 2}}=
\end{aligned}
$$

$$
\begin{aligned}
& =\frac{\sqrt{(\sqrt{3}-\sqrt{2})^{2}} \cdot(\sqrt{3}+\sqrt{2})^{2}(49-20 \sqrt{6})}{3 \sqrt{3}-9 \sqrt{2}+6 \sqrt{3}-2 \sqrt{2}}=\frac{(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})^{2}(49-20 \sqrt{6})}{9 \sqrt{3}-11 \sqrt{2}}= \\
& =\frac{((\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})) \cdot(\sqrt{3}+\sqrt{2})(49-20 \sqrt{6})}{9 \sqrt{3}-11 \sqrt{2}}= \\
& =\frac{\left((\sqrt{3})^{2}-(\sqrt{2})^{2}\right) \cdot(49 \sqrt{3}-20 \sqrt{18}+49 \sqrt{2}-20 \sqrt{12})}{9 \sqrt{3}-11 \sqrt{2}}= \\
& =\frac{(3-2)(49 \sqrt{3}-20 \sqrt{9 \cdot 2}+49 \sqrt{2}-20 \sqrt{4 \cdot 3})}{9 \sqrt{3}-11 \sqrt{2}}= \\
& =\frac{49 \sqrt{3}-60 \sqrt{2}+49 \sqrt{2}-40 \sqrt{3}}{9 \sqrt{3}-11 \sqrt{2}}=\frac{9 \sqrt{3}-11 \sqrt{2}}{9 \sqrt{3}-11 \sqrt{2}}=1 .
\end{aligned}
$$