We are tasked with rationalizing the denominator of:

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

Step 1: Multiply by the conjugate

To rationalize the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of $5\sqrt{6} - \sqrt{5}$ is $5\sqrt{6} + \sqrt{5}$. This works because the product of a binomial and its conjugate is a difference of squares.

Thus, multiply the expression by:

$\frac{5\sqrt{6} + \sqrt{5}}{5\sqrt{6} + \sqrt{5}}$

This gives:

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

Step 2: Simplify the denominator

The denominator becomes the difference of squares:

$(5\sqrt{6})^2 - (\sqrt{5})^2 = 25 \times 6 - 5 = 150 - 5 = 145$

Step 3: Simplify the numerator

Distribute $\sqrt{5}$ in the numerator:

$\sqrt{5}(5\sqrt{6}) + \sqrt{5}(\sqrt{5}) = 5\sqrt{30} + 5$

Step 4: Write the final expression

Now, the expression becomes:

$\frac{5\sqrt{30} + 5}{145}$

This can be simplified further as:

$\frac{5(\sqrt{30} + 1)}{145}$

Since there is no further simplification, this is the rationalized form.

Final Answer:

$\frac{5(\sqrt{30} + 1)}{145}$

Would you like more details on any step or do you have questions?

Here are 5 related questions you might find helpful:

1. How do you rationalize more complex expressions with multiple square roots?
2. What is the benefit of rationalizing the denominator in mathematical expressions?
3. How do difference of squares formulas help in simplifying radicals?
4. Can you apply this technique to rationalize complex fractions?
5. How would this change if the denominator involved cube roots?

Tip: Always check if a binomial has a conjugate; multiplying by it simplifies many radical denominators.