To rationalize the denominator of the expression $\frac{1}{\sqrt{7} - \sqrt{6}}$, we need to multiply both the numerator and denominator by the conjugate of the denominator. The conjugate of $\sqrt{7} - \sqrt{6}$ is $\sqrt{7} + \sqrt{6}$. This process eliminates the square roots in the denominator.

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:

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

So, the expression simplifies to:

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

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

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

Would you like further clarification or examples?

Here are some related questions to explore:

1. How do you rationalize denominators with binomials involving addition or subtraction?
2. What is the difference between rationalizing a denominator and simplifying an expression?
3. Can you explain the difference of squares formula and how it's used in rationalizing denominators?
4. How would you rationalize an expression with a cube root in the denominator?
5. What happens if the numerator also has a square root term when rationalizing the denominator?

Tip: When rationalizing denominators with square roots, always use the conjugate (the binomial formed by changing the sign between the terms) to simplify the expression.