We are asked to rationalize the denominator and simplify the given expressions: a. $\frac{3}{\sqrt{2}}$ b. $\frac{3}{2-\sqrt{3}}$ c. $\frac{3-\sqrt{5}}{1+3\sqrt{5}}$

1. Problem Description

We are asked to rationalize the denominator and simplify the given expressions:

a. $\frac{3}{\sqrt{2}}$

b. $\frac{3}{2-\sqrt{3}}$

c. $\frac{3-\sqrt{5}}{1+3\sqrt{5}}$

2. Solution Steps

a. To rationalize the denominator of $\frac{3}{\sqrt{2}}$, we multiply both the numerator and denominator by $\sqrt{2}$:

$\frac{3}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{3\sqrt{2}}{2}$

b. To rationalize the denominator of $\frac{3}{2-\sqrt{3}}$, we multiply both the numerator and denominator by the conjugate of the denominator, which is $2+\sqrt{3}$:

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

The denominator becomes $(2-\sqrt{3})(2+\sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4 - 3 = 1$.

Therefore, $\frac{3(2+\sqrt{3})}{1} = 6 + 3\sqrt{3}$

c. To rationalize the denominator of $\frac{3-\sqrt{5}}{1+3\sqrt{5}}$, we multiply both the numerator and denominator by the conjugate of the denominator, which is $1-3\sqrt{5}$:

$\frac{3-\sqrt{5}}{1+3\sqrt{5}} \cdot \frac{1-3\sqrt{5}}{1-3\sqrt{5}} = \frac{(3-\sqrt{5})(1-3\sqrt{5})}{(1+3\sqrt{5})(1-3\sqrt{5})}$

The numerator is $(3-\sqrt{5})(1-3\sqrt{5}) = 3 - 9\sqrt{5} - \sqrt{5} + 3(\sqrt{5})^2 = 3 - 10\sqrt{5} + 3(5) = 3 - 10\sqrt{5} + 15 = 18 - 10\sqrt{5}$.

The denominator is $(1+3\sqrt{5})(1-3\sqrt{5}) = 1^2 - (3\sqrt{5})^2 = 1 - 9(5) = 1 - 45 = -44$.

Therefore, $\frac{18 - 10\sqrt{5}}{-44} = \frac{2(9 - 5\sqrt{5})}{2(-22)} = \frac{9 - 5\sqrt{5}}{-22} = \frac{-9 + 5\sqrt{5}}{22}$.

3. Final Answer

a. $\frac{3\sqrt{2}}{2}$

b. $6 + 3\sqrt{3}$

c. $\frac{5\sqrt{5}-9}{22}$