We are asked to solve three problems: e) $\frac{3}{1 + \sqrt{3}}$ f) $\frac{\sqrt{2}+1}{\sqrt{2}-1}$ g) $\frac{2\sqrt{5}}{\sqrt{2}-\sqrt{3}}$

1. Problem Description

We are asked to solve three problems:

e) $\frac{3}{1 + \sqrt{3}}$

f) $\frac{\sqrt{2}+1}{\sqrt{2}-1}$

g) $\frac{2\sqrt{5}}{\sqrt{2}-\sqrt{3}}$

2. Solution Steps

e) To simplify $\frac{3}{1 + \sqrt{3}}$, we multiply the numerator and denominator by the conjugate of the denominator, which is $1 - \sqrt{3}$.

$\frac{3}{1 + \sqrt{3}} = \frac{3(1 - \sqrt{3})}{(1 + \sqrt{3})(1 - \sqrt{3})} = \frac{3(1 - \sqrt{3})}{1 - (\sqrt{3})^2} = \frac{3(1 - \sqrt{3})}{1 - 3} = \frac{3(1 - \sqrt{3})}{-2} = \frac{3(\sqrt{3} - 1)}{2}$

f) To simplify $\frac{\sqrt{2}+1}{\sqrt{2}-1}$, we multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{2} + 1$.

$\frac{\sqrt{2}+1}{\sqrt{2}-1} = \frac{(\sqrt{2}+1)(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)} = \frac{(\sqrt{2}+1)^2}{(\sqrt{2})^2 - 1^2} = \frac{2 + 2\sqrt{2} + 1}{2 - 1} = \frac{3 + 2\sqrt{2}}{1} = 3 + 2\sqrt{2}$

g) To simplify $\frac{2\sqrt{5}}{\sqrt{2}-\sqrt{3}}$, we multiply the numerator and denominator by the conjugate of the denominator, which is $\sqrt{2} + \sqrt{3}$.

$\frac{2\sqrt{5}}{\sqrt{2}-\sqrt{3}} = \frac{2\sqrt{5}(\sqrt{2} + \sqrt{3})}{(\sqrt{2}-\sqrt{3})(\sqrt{2}+\sqrt{3})} = \frac{2\sqrt{5}(\sqrt{2} + \sqrt{3})}{(\sqrt{2})^2 - (\sqrt{3})^2} = \frac{2(\sqrt{10} + \sqrt{15})}{2 - 3} = \frac{2(\sqrt{10} + \sqrt{15})}{-1} = -2(\sqrt{10} + \sqrt{15})$

3. Final Answer

e) $\frac{3(\sqrt{3} - 1)}{2}$

f) $3 + 2\sqrt{2}$

g) $-2(\sqrt{10} + \sqrt{15})$