2.150. $\frac{2-\sqrt{2}-\sqrt{3}}{2+\sqrt{2}-\sqrt{3}}$.
## Solution.

Let's represent the given fraction as $\frac{\sqrt{4}-\sqrt{2}-\sqrt{3}}{\sqrt{4}+\sqrt{2}-\sqrt{3}} \cdot$ Multiply this fraction by $(\sqrt{4}+\sqrt{2}+\sqrt{3})(4+2-3-2 \sqrt{4 \cdot 2})$ and, applying the equality $(\sqrt{a}+\sqrt{b}-\sqrt{c})(\sqrt{a}+\sqrt{b}-\sqrt{c})(a+b-c-2 \sqrt{a b})=(a+b+c)^{2}-4 a b, \quad$ where $a \geq 0, b \geq 0$ and $c \geq 0$, we get

$$
\frac{(\sqrt{4}-\sqrt{2}-\sqrt{3})(\sqrt{4}+\sqrt{2}+\sqrt{3})(4+2-3-2 \sqrt{4 \cdot 2})}{(\sqrt{4}+\sqrt{2}-\sqrt{3})(\sqrt{4}+\sqrt{2}+\sqrt{3})(4+2-3-2 \sqrt{4 \cdot 2})}=
$$

$=\frac{\left((\sqrt{4})^{2}-(\sqrt{2}+\sqrt{3})^{2}\right)(3-4 \sqrt{2})}{(4+2-3)^{2}-4 \cdot 4 \cdot 2}=\frac{(4-(2+2 \sqrt{6}+3)(3-4 \sqrt{2})}{9-32}=$

$=\frac{(2 \sqrt{6}+1)(3-4 \sqrt{2})}{23}$.

Answer: $\frac{(2 \sqrt{6}+1)(3-4 \sqrt{2})}{23}$.