The problem asks us to simplify the expression $\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}$ by rationalizing the denominator.

1. Problem Description

The problem asks us to simplify the expression $\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}}$ by rationalizing the denominator.

2. Solution Steps

To rationalize the denominator, we multiply the numerator and denominator by the conjugate of the denominator. The conjugate of $\sqrt{7}-\sqrt{3}$ is $\sqrt{7}+\sqrt{3}$.

$\frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}-\sqrt{3}} \cdot \frac{\sqrt{7}+\sqrt{3}}{\sqrt{7}+\sqrt{3}}$

Now, we multiply the numerators and the denominators:

$\frac{(\sqrt{7}+\sqrt{3})(\sqrt{7}+\sqrt{3})}{(\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3})}$

Expanding the numerator, we get:

$(\sqrt{7}+\sqrt{3})(\sqrt{7}+\sqrt{3}) = (\sqrt{7})^2 + 2(\sqrt{7})(\sqrt{3}) + (\sqrt{3})^2 = 7 + 2\sqrt{21} + 3 = 10 + 2\sqrt{21}$

Expanding the denominator, we use the difference of squares formula $(a-b)(a+b) = a^2 - b^2$:

$(\sqrt{7}-\sqrt{3})(\sqrt{7}+\sqrt{3}) = (\sqrt{7})^2 - (\sqrt{3})^2 = 7 - 3 = 4$

So the expression becomes:

$\frac{10 + 2\sqrt{21}}{4}$

Now, we can simplify by dividing both terms in the numerator by 2:

$\frac{10 + 2\sqrt{21}}{4} = \frac{2(5 + \sqrt{21})}{4} = \frac{5 + \sqrt{21}}{2}$

3. Final Answer

$\frac{5 + \sqrt{21}}{2}$