To simplify the expression given in the image:

$\frac{\sqrt{8} - \sqrt{6}}{\sqrt{8} + \sqrt{6}}$

using the method of rationalizing the denominator, we multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of $\sqrt{8} + \sqrt{6}$ is $\sqrt{8} - \sqrt{6}$.

So, multiply both the numerator and denominator by $\sqrt{8} - \sqrt{6}$:

$\frac{\sqrt{8} - \sqrt{6}}{\sqrt{8} + \sqrt{6}} \times \frac{\sqrt{8} - \sqrt{6}}{\sqrt{8} - \sqrt{6}}$

This gives:

$\frac{(\sqrt{8} - \sqrt{6})^2}{(\sqrt{8})^2 - (\sqrt{6})^2}$

Now simplify:

1. Numerator: $(\sqrt{8} - \sqrt{6})^2 = (\sqrt{8})^2 - 2\sqrt{8}\sqrt{6} + (\sqrt{6})^2 = 8 - 2\sqrt{48} + 6 = 14 - 2\sqrt{48}$

Since $\sqrt{48} = 4\sqrt{3}$, the numerator becomes: $14 - 8\sqrt{3}$

2. Denominator: $(\sqrt{8})^2 - (\sqrt{6})^2 = 8 - 6 = 2$

Now, the expression becomes: $\frac{14 - 8\sqrt{3}}{2}$

Finally, simplify by dividing both terms in the numerator by 2: $7 - 4\sqrt{3}$

Thus, the simplified expression is:

$7 - 4\sqrt{3}$

Would you like a detailed explanation of any of the steps?

Here are five related questions for further practice:

1. How do you simplify $\frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$ using rationalization?
2. What is the conjugate of $\sqrt{a} + \sqrt{b}$, and why is it used in rationalization?
3. How would you rationalize $\frac{1}{\sqrt{2} + \sqrt{3}}$?
4. What is the simplified result of $\frac{\sqrt{10} - \sqrt{2}}{\sqrt{10} + \sqrt{2}}$?
5. Can you apply rationalization to a denominator of the form $a + b\sqrt{c}$? How?

Tip: When rationalizing, always multiply by the conjugate of the denominator to eliminate radicals.