The question is asking to simplify the expression to its rational form:

$\frac{6\sqrt{2}}{\sqrt{7} + 2\sqrt{2}}$

To rationalize the denominator, we will multiply both the numerator and the denominator by the conjugate of the denominator, which is $\sqrt{7} - 2\sqrt{2}$.

$\text{Multiply the expression by } \frac{\sqrt{7} - 2\sqrt{2}}{\sqrt{7} - 2\sqrt{2}}:$

$\frac{6\sqrt{2}(\sqrt{7} - 2\sqrt{2})}{(\sqrt{7} + 2\sqrt{2})(\sqrt{7} - 2\sqrt{2})}$

Now we simplify the denominator:

$(\sqrt{7} + 2\sqrt{2})(\sqrt{7} - 2\sqrt{2}) = (\sqrt{7})^2 - (2\sqrt{2})^2 = 7 - 8 = -1$

Now simplify the numerator:

$6\sqrt{2}(\sqrt{7} - 2\sqrt{2}) = 6\sqrt{14} - 12\sqrt{4} = 6\sqrt{14} - 12 \times 2 = 6\sqrt{14} - 24$

Putting it all together:

$\frac{6\sqrt{14} - 24}{-1} = -(6\sqrt{14} - 24) = -6\sqrt{14} + 24$

Thus, the rationalized form is:

$24 - 6\sqrt{14}$

Would you like a detailed explanation of any specific steps?

Here are 5 related questions to expand your understanding:

1. How do you rationalize a denominator with two terms involving square roots?
2. What is the significance of multiplying by the conjugate in rationalizing denominators?
3. Can this technique of rationalization be applied to complex numbers?
4. How do you simplify expressions with square roots, such as $\sqrt{14}$?
5. Why do we simplify terms like $\sqrt{2} \times \sqrt{2} = 2$?

Tip: Always look for conjugates when rationalizing a denominator with two terms involving square roots.