To rationalize the denominator of the expression $\frac{1}{\sqrt{7} - \sqrt{6}}$, we multiply both the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{7} + \sqrt{6}$.

Here are the steps:

$\frac{1}{\sqrt{7} - \sqrt{6}} \times \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} = \frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7} - \sqrt{6})(\sqrt{7} + \sqrt{6})}$

Now, simplify the denominator using the difference of squares formula $(a - b)(a + b) = a^2 - b^2$:

$(\sqrt{7})^2 - (\sqrt{6})^2 = 7 - 6 = 1$

Thus, the expression becomes:

$\frac{\sqrt{7} + \sqrt{6}}{1} = \sqrt{7} + \sqrt{6}$

So, the rationalized form of $\frac{1}{\sqrt{7} - \sqrt{6}}$ is:

$\boxed{\sqrt{7} + \sqrt{6}}$

Would you like a further explanation of any of the steps, or have any other questions on this? Here's some related questions for you:

1. How do you rationalize denominators with binomial expressions involving square roots?
2. What happens if we multiply the numerator and denominator by a term other than the conjugate?
3. Can you explain the difference of squares formula and provide another example?
4. How would you rationalize the denominator of an expression with cube roots?
5. Why is rationalizing the denominator important in mathematics?

Tip: Always check if the denominator is a binomial (two terms) involving square roots or other radicals—this is when using the conjugate is most effective.