Simplify the expression $\frac{\sqrt{3}}{\sqrt{3} + \sqrt{2}}$.

1. Problem Description

Simplify the expression $\frac{\sqrt{3}}{\sqrt{3} + \sqrt{2}}$.

2. Solution Steps

To simplify the expression, we need to rationalize the denominator. We can do this by multiplying the numerator and denominator by the conjugate of the denominator.

The conjugate of $\sqrt{3} + \sqrt{2}$ is $\sqrt{3} - \sqrt{2}$.

Multiply both the numerator and the denominator by $\sqrt{3} - \sqrt{2}$:

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

Now multiply the numerators:

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

And multiply the denominators:

$(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$

So the expression becomes:

$\frac{3 - \sqrt{6}}{1} = 3 - \sqrt{6}$

3. Final Answer

$3 - \sqrt{6}$