The problem asks us to simplify the expression $\frac{x}{\sqrt{13}-\sqrt{7}}$ by rationalizing the denominator. This means we need to eliminate the radicals from the denominator.

1. Problem Description

The problem asks us to simplify the expression $\frac{x}{\sqrt{13}-\sqrt{7}}$ by rationalizing the denominator. This means we need to eliminate the radicals from the denominator.

2. Solution Steps

To rationalize the denominator, we multiply the numerator and the denominator by the conjugate of the denominator. The conjugate of $\sqrt{13}-\sqrt{7}$ is $\sqrt{13}+\sqrt{7}$.

So we have:

$\frac{x}{\sqrt{13}-\sqrt{7}} = \frac{x}{\sqrt{13}-\sqrt{7}} \cdot \frac{\sqrt{13}+\sqrt{7}}{\sqrt{13}+\sqrt{7}}$

$= \frac{x(\sqrt{13}+\sqrt{7})}{(\sqrt{13}-\sqrt{7})(\sqrt{13}+\sqrt{7})}$

Using the difference of squares formula $(a-b)(a+b) = a^2 - b^2$, we have:

$(\sqrt{13}-\sqrt{7})(\sqrt{13}+\sqrt{7}) = (\sqrt{13})^2 - (\sqrt{7})^2 = 13 - 7 = 6$

Therefore, the expression becomes:

$\frac{x(\sqrt{13}+\sqrt{7})}{6} = \frac{x\sqrt{13}+x\sqrt{7}}{6}$

3. Final Answer

$\frac{x(\sqrt{13}+\sqrt{7})}{6}$