1. **State the problem:** Simplify the expression $$\frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}}$$ into the form $$a + \sqrt{b}$$ and find the values of $$a$$ and $$b$$. 2. **Formula and rule:** To simplify expressions with radicals in the denominator, 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:** $$\frac{\sqrt{2}}{\sqrt{3}+\sqrt{2}} \times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}} = \frac{\sqrt{2}(\sqrt{3}-\sqrt{2})}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$$ 4. **Simplify the denominator using difference of squares:** $$(\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1$$ 5. **Simplify the numerator:** $$\sqrt{2} \times \sqrt{3} - \sqrt{2} \times \sqrt{2} = \sqrt{6} - 2$$ 6. **Put it all together:** $$\frac{\sqrt{6} - 2}{1} = \sqrt{6} - 2$$ 7. **Rewrite in the form $$a + \sqrt{b}$$:** $$a = -2, \quad b = 6$$ **Final answer:** $$a = -2$$ and $$b = 6$$.