The question asks to rationalize the denominator of the expression $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}$.

Solution:

To rationalize the denominator, multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of $\sqrt{5} - \sqrt{3}$ is $\sqrt{5} + \sqrt{3}$.

So, we multiply the fraction by $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}}$:

$\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} \times \frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} + \sqrt{3}} = \frac{(\sqrt{5} + \sqrt{3})^2}{(\sqrt{5})^2 - (\sqrt{3})^2}$

Step 1: Simplify the numerator

$(\sqrt{5} + \sqrt{3})^2 = (\sqrt{5})^2 + 2(\sqrt{5})(\sqrt{3}) + (\sqrt{3})^2 = 5 + 2\sqrt{15} + 3 = 8 + 2\sqrt{15}$

Step 2: Simplify the denominator

$(\sqrt{5})^2 - (\sqrt{3})^2 = 5 - 3 = 2$

Step 3: Combine the results

$\frac{8 + 2\sqrt{15}}{2}$

Step 4: Simplify the fraction

$= \frac{8}{2} + \frac{2\sqrt{15}}{2} = 4 + \sqrt{15}$

So, the rationalized expression is $4 + \sqrt{15}$.

Would you like any more details or have any other questions?

Here are five related questions for further practice:

1. Rationalize the denominator of $\frac{\sqrt{7} + \sqrt{2}}{\sqrt{7} - \sqrt{2}}$.
2. Simplify $\frac{3\sqrt{2} + 2\sqrt{3}}{2\sqrt{2} - \sqrt{3}}$.
3. Rationalize the denominator and simplify $\frac{1 + \sqrt{5}}{\sqrt{5} - 1}$.
4. Find the value of $\frac{\sqrt{6} + \sqrt{2}}{\sqrt{6} - \sqrt{2}}$.
5. Simplify $\frac{5\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$.

Tip: When rationalizing denominators, always remember to multiply by the conjugate of the denominator, which changes the sign between the two terms.