1. **State the problem:** Rationalize the expression $$\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}}$$. 2. **Formula and rule:** To rationalize a denominator with radicals, multiply numerator and denominator by the conjugate of the denominator. The conjugate of $$\sqrt{3} - \sqrt{2}$$ is $$\sqrt{3} + \sqrt{2}$$. 3. **Multiply numerator and denominator by the conjugate:** $$\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \times \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} + \sqrt{2}} = \frac{(\sqrt{3} + \sqrt{2})^2}{(\sqrt{3})^2 - (\sqrt{2})^2}$$ 4. **Simplify numerator:** $$(\sqrt{3} + \sqrt{2})^2 = (\sqrt{3})^2 + 2\sqrt{3}\sqrt{2} + (\sqrt{2})^2 = 3 + 2\sqrt{6} + 2 = 5 + 2\sqrt{6}$$ 5. **Simplify denominator:** $$(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$$ 6. **Final expression:** $$\frac{5 + 2\sqrt{6}}{1} = 5 + 2\sqrt{6}$$ **Answer:** $$5 + 2\sqrt{6}$$