https://file.aiquickdraw.com/mathbot/file/1727310623108eienwkpq.jpg

Let's rationalize the denominator for each expression:

### Part a) \( \frac{6}{2\sqrt{7} - 5} \)

To rationalize the denominator, we multiply both the numerator and denominator by the conjugate of \( 2\sqrt{7} - 5 \), which is \( 2\sqrt{7} + 5 \).

\[
\frac{6}{2\sqrt{7} - 5} \times \frac{2\sqrt{7} + 5}{2\sqrt{7} + 5} = \frac{6(2\sqrt{7} + 5)}{(2\sqrt{7})^2 - 5^2}
\]

Simplifying the denominator:

\[
(2\sqrt{7})^2 = 4 \times 7 = 28,\quad 5^2 = 25
\]

So the denominator becomes \( 28 - 25 = 3 \).

Now simplifying the entire expression:

\[
\frac{6(2\sqrt{7} + 5)}{3} = 2(2\sqrt{7} + 5)
\]

Therefore, the result is:

\[
4\sqrt{7} + 10
\]

---

### Part b) \( \frac{2}{\sqrt{5} - 1} \)

Multiply both the numerator and denominator by the conjugate of \( \sqrt{5} - 1 \), which is \( \sqrt{5} + 1 \).

\[
\frac{2}{\sqrt{5} - 1} \times \frac{\sqrt{5} + 1}{\sqrt{5} + 1} = \frac{2(\sqrt{5} + 1)}{(\sqrt{5})^2 - 1^2}
\]

Simplifying the denominator:

\[
(\sqrt{5})^2 = 5,\quad 1^2 = 1
\]

So the denominator becomes \( 5 - 1 = 4 \).

Now simplifying the entire expression:

\[
\frac{2(\sqrt{5} + 1)}{4} = \frac{\sqrt{5} + 1}{2}
\]

---

### Part c) \( \frac{\sqrt{3} + 5}{\sqrt{3} - 1} \)

Again, we multiply both the numerator and denominator by the conjugate of \( \sqrt{3} - 1 \), which is \( \sqrt{3} + 1 \).

\[
\frac{\sqrt{3} + 5}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} + 5)(\sqrt{3} + 1)}{(\sqrt{3})^2 - 1^2}
\]

Simplifying the denominator:

\[
(\sqrt{3})^2 = 3,\quad 1^2 = 1
\]

So the denominator becomes \( 3 - 1 = 2 \).

Now expanding the numerator:

\[
(\sqrt{3} + 5)(\sqrt{3} + 1) = \sqrt{3} \times \sqrt{3} + \sqrt{3} \times 1 + 5 \times \sqrt{3} + 5 \times 1
= 3 + \sqrt{3} + 5\sqrt{3} + 5
= 3 + 5 + \sqrt{3}(1 + 5) = 8 + 6\sqrt{3}
\]

Now divide the numerator by 2:

\[
\frac{8 + 6\sqrt{3}}{2} = 4 + 3\sqrt{3}
\]

---

### Final Answers:

- **a)** \( 4\sqrt{7} + 10 \)
- **b)** \( \frac{\sqrt{5} + 1}{2} \)
- **c)** \( 4 + 3\sqrt{3} \)

Would you like more details on any of these steps or further clarifications?

### Relative Questions:
1. How do you determine the conjugate of a binomial with a square root?
2. Why does multiplying by the conjugate help in rationalizing denominators?
3. What would happen if you tried rationalizing with an incorrect conjugate?
4. Can you rationalize the denominator if the binomial contains higher powers of square roots?
5. How does rationalization affect the overall expression?

### Tip:
Always check if the denominator contains an expression of the form \( a\sqrt{b} - c \) and multiply by its conjugate \( a\sqrt{b} + c \) to easily eliminate the square roots.

Rationalize the denominator for each expression:
 a) \( \frac{6}{2\sqrt{7} - 5} \) 
 b) \( \frac{2}{\sqrt{5} - 1} \) 
 c) \( \frac{\sqrt{3} + 5}{\sqrt{3} - 1} \)

Learn how to rationalize the denominator for expressions involving square roots and binomials, including detailed solutions for expressions like \( \frac{6}{2\sqrt{7} - 5} \), \( \frac{2}{\sqrt{5} - 1} \), and \( \frac{\sqrt{3} + 5}{\sqrt{3} - 1} \). Suitable for students in Grades 10-12.

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