To rationalize the numerator of the given expression:

$\frac{6 - \sqrt{3}}{5}$

we multiply both the numerator and the denominator by the conjugate of the numerator, which is $6 + \sqrt{3}$, to eliminate the square root from the numerator. The expression becomes:

$\frac{(6 - \sqrt{3})(6 + \sqrt{3})}{5(6 + \sqrt{3})}$

Now, let's simplify this step by step:

1. Apply the difference of squares formula: $(a - b)(a + b) = a^2 - b^2$.

$(6 - \sqrt{3})(6 + \sqrt{3}) = 6^2 - (\sqrt{3})^2 = 36 - 3 = 33$

2. The denominator becomes:

$5(6 + \sqrt{3}) = 5(6 + \sqrt{3})$

Thus, the rationalized form of the expression is:

$\frac{33}{5(6 + \sqrt{3})}$

This is the final rationalized form of the expression. Would you like more details on any of these steps?

Here are five questions to expand on this:

1. What is the conjugate of a binomial containing a square root?
2. How does the difference of squares formula help in rationalization?
3. Can you rationalize the denominator in this form?
4. What are the steps to simplify an expression involving radicals?
5. How would you rationalize a more complex expression like $\frac{a - \sqrt{b}}{c}$?

Tip: Always multiply by the conjugate when rationalizing an expression with a square root in the numerator or denominator.