The given mathematical problem in the image is:

$\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{\sqrt{2} + \sqrt{3} + \sqrt{5}} = ?$

Let us solve this step by step:

Step 1: Rationalize the denominator

To simplify the given expression, multiply the numerator and the denominator by the conjugate of the denominator $(\sqrt{2} + \sqrt{3} - \sqrt{5})$. The conjugate is obtained by flipping the sign of the last term of the denominator.

$\text{Conjugate of the denominator: } \sqrt{2} + \sqrt{3} - \sqrt{5}$

Thus, the expression becomes:

$\frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{\sqrt{2} + \sqrt{3} + \sqrt{5}} \cdot \frac{\sqrt{2} + \sqrt{3} - \sqrt{5}}{\sqrt{2} + \sqrt{3} - \sqrt{5}}$

Step 2: Simplify the denominator

The denominator is of the form $(a + b)(a - b)$, where $a = \sqrt{2} + \sqrt{3}$ and $b = \sqrt{5}$. Using the difference of squares formula:

$(a + b)(a - b) = a^2 - b^2$

Here: $a^2 = (\sqrt{2} + \sqrt{3})^2 = (\sqrt{2})^2 + 2(\sqrt{2})(\sqrt{3}) + (\sqrt{3})^2 = 2 + 2\sqrt{6} + 3 = 5 + 2\sqrt{6}$

$b^2 = (\sqrt{5})^2 = 5$

So: $\text{Denominator: } a^2 - b^2 = (5 + 2\sqrt{6}) - 5 = 2\sqrt{6}$

Step 3: Simplify the numerator

The numerator is $(\sqrt{2} + \sqrt{3} - \sqrt{5})^2$. Expanding this:

$(\sqrt{2} + \sqrt{3} - \sqrt{5})^2 = (\sqrt{2})^2 + (\sqrt{3})^2 + (\sqrt{5})^2 + 2(\sqrt{2})(\sqrt{3}) - 2(\sqrt{2})(\sqrt{5}) - 2(\sqrt{3})(\sqrt{5})$

$= 2 + 3 + 5 + 2\sqrt{6} - 2\sqrt{10} - 2\sqrt{15}$

$= 10 + 2\sqrt{6} - 2\sqrt{10} - 2\sqrt{15}$

Step 4: Write the simplified fraction

Now, the fraction becomes:

$\frac{10 + 2\sqrt{6} - 2\sqrt{10} - 2\sqrt{15}}{2\sqrt{6}}$

Step 5: Simplify each term

Divide each term in the numerator by the denominator ($2\sqrt{6}$):

$\frac{10}{2\sqrt{6}} + \frac{2\sqrt{6}}{2\sqrt{6}} - \frac{2\sqrt{10}}{2\sqrt{6}} - \frac{2\sqrt{15}}{2\sqrt{6}}$

Simplify each term:

1. $\frac{10}{2\sqrt{6}} = \frac{5}{\sqrt{6}} = \frac{5\sqrt{6}}{6}$ (rationalizing the denominator),
2. $\frac{2\sqrt{6}}{2\sqrt{6}} = 1$,
3. $\frac{2\sqrt{10}}{2\sqrt{6}} = \frac{\sqrt{10}}{\sqrt{6}} = \frac{\sqrt{60}}{6} = \frac{2\sqrt{15}}{6} = \frac{\sqrt{15}}{3}$,
4. $\frac{2\sqrt{15}}{2\sqrt{6}} = \frac{\sqrt{15}}{\sqrt{6}} = \frac{\sqrt{90}}{6} = \frac{3\sqrt{10}}{6} = \frac{\sqrt{10}}{2}$.

Step 6: Combine terms

The final result is:

$\frac{5\sqrt{6}}{6} + 1 - \frac{\sqrt{15}}{3} - \frac{\sqrt{10}}{2}$

Let me know if you would like further clarification or to simplify this result even further!

Here are 5 related questions for practice:

1. How do you rationalize the denominator of a fraction involving square roots?
2. What is the formula for expanding $(a + b - c)^2$?
3. Can you prove the difference of squares formula for radicals?
4. How would you simplify $\frac{\sqrt{a} + \sqrt{b}}{\sqrt{c}}$?
5. How can the concept of conjugates be applied to simplify irrational denominators?

Tip: When simplifying square root expressions, always try to rationalize denominators to make the final result cleaner and more interpretable.