To solve the given expression, we need to simplify the expression by rationalizing the denominators. Here is the step-by-step solution:

$\frac{3}{\sqrt{2} - 2} - \frac{2}{\sqrt{2} + 1}$

Step 1: Rationalize the Denominators

For $\frac{3}{\sqrt{2} - 2}$:

Multiply the numerator and the denominator by the conjugate of the denominator, $\sqrt{2} + 2$:

$\frac{3}{\sqrt{2} - 2} \times \frac{\sqrt{2} + 2}{\sqrt{2} + 2} = \frac{3(\sqrt{2} + 2)}{(\sqrt{2} - 2)(\sqrt{2} + 2)}$

Simplify the denominator:

$(\sqrt{2} - 2)(\sqrt{2} + 2) = (\sqrt{2})^2 - (2)^2 = 2 - 4 = -2$

So:

$\frac{3(\sqrt{2} + 2)}{-2} = \frac{3\sqrt{2} + 6}{-2} = -\frac{3\sqrt{2}}{2} - 3$

For $\frac{2}{\sqrt{2} + 1}$:

Multiply the numerator and the denominator by the conjugate of the denominator, $\sqrt{2} - 1$:

$\frac{2}{\sqrt{2} + 1} \times \frac{\sqrt{2} - 1}{\sqrt{2} - 1} = \frac{2(\sqrt{2} - 1)}{(\sqrt{2} + 1)(\sqrt{2} - 1)}$

Simplify the denominator:

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

So:

$\frac{2(\sqrt{2} - 1)}{1} = 2\sqrt{2} - 2$

Step 2: Combine the Two Expressions

Now we combine the two simplified expressions:

$-\frac{3\sqrt{2}}{2} - 3 - (2\sqrt{2} - 2)$

Distribute the negative sign and combine like terms:

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

Convert $2\sqrt{2}$ to a common denominator:

$2\sqrt{2} = \frac{4\sqrt{2}}{2}$

Combine like terms:

$-\frac{3\sqrt{2}}{2} - \frac{4\sqrt{2}}{2} - 3 + 2 = -\frac{7\sqrt{2}}{2} - 1$

So, the simplified expression is:

$-\frac{7\sqrt{2}}{2} - 1$

Would you like more details or have any questions about this solution? Here are five questions you might ask next:

1. How do you rationalize the denominator of a fraction?
2. Can you explain the process of multiplying by the conjugate in more detail?
3. How do you simplify expressions involving square roots?
4. Why is it necessary to combine like terms in algebraic expressions?
5. Can you show another example of simplifying a similar expression?

Tip: When dealing with square roots and fractions, rationalizing the denominator can often simplify the problem significantly.