To solve the expression $\frac{2 + \sqrt{3}}{\sqrt{3} - \sqrt{2}}$, we can start by rationalizing the denominator. This involves multiplying both the numerator and the denominator by the conjugate of the denominator.

Steps:

1. Identify the conjugate of the denominator:
   The conjugate of $\sqrt{3} - \sqrt{2}$ is $\sqrt{3} + \sqrt{2}$.

2. Multiply both numerator and denominator by the conjugate: $\frac{2 + \sqrt{3}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{(2 + \sqrt{3})(\sqrt{3} + \sqrt{2})}{(\sqrt{3} - \sqrt{2})(\sqrt{3} + \sqrt{2})}$

3. Expand the numerator: $(2 + \sqrt{3})(\sqrt{3} + \sqrt{2}) = 2\sqrt{3} + 2\sqrt{2} + 3 + \sqrt{6}$ Simplified, this is: $3 + 2\sqrt{2} + 2\sqrt{3} + \sqrt{6}$

4. Simplify the denominator: The denominator simplifies as the difference of squares: $(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$

5. Final Expression: Since the denominator is 1, the final expression is just the simplified numerator: $3 + 2\sqrt{2} + 2\sqrt{3} + \sqrt{6}$

So, the simplified form of $\frac{2 + \sqrt{3}}{\sqrt{3} - \sqrt{2}}$ is: $\boxed{3 + 2\sqrt{2} + 2\sqrt{3} + \sqrt{6}}$

Would you like further details or have any questions about this process?

Related Questions:

1. How do you rationalize more complex denominators?
2. What is the significance of the conjugate in rationalizing denominators?
3. Can you explain how the difference of squares is used in rationalizing?
4. How does rationalization help in solving real-life problems?
5. Can you provide an example where rationalization simplifies the problem-solving process?

Tip:

When simplifying expressions involving square roots, always check if rationalizing the denominator can make the expression easier to work with.