1. **State the problem:** Simplify the expression $$\left(\frac{\sqrt{2}-1}{2}\right)^{-1} - \left(\frac{1}{\sqrt{2}+1}\right)$$. 2. **Recall the rule for negative exponents:** For any nonzero number $a$, $a^{-1} = \frac{1}{a}$. 3. **Apply the negative exponent:** $$\left(\frac{\sqrt{2}-1}{2}\right)^{-1} = \frac{1}{\frac{\sqrt{2}-1}{2}} = \frac{2}{\sqrt{2}-1}$$ 4. **Rewrite the expression:** $$\frac{2}{\sqrt{2}-1} - \frac{1}{\sqrt{2}+1}$$ 5. **Rationalize the denominators:** - For the first term: $$\frac{2}{\sqrt{2}-1} \times \frac{\sqrt{2}+1}{\sqrt{2}+1} = \frac{2(\sqrt{2}+1)}{(\sqrt{2})^2 - 1^2} = \frac{2(\sqrt{2}+1)}{2 - 1} = 2(\sqrt{2}+1)$$ - For the second term: $$\frac{1}{\sqrt{2}+1} \times \frac{\sqrt{2}-1}{\sqrt{2}-1} = \frac{\sqrt{2}-1}{(\sqrt{2})^2 - 1^2} = \frac{\sqrt{2}-1}{2 - 1} = \sqrt{2} - 1$$ 6. **Substitute back:** $$2(\sqrt{2}+1) - (\sqrt{2} - 1)$$ 7. **Distribute and simplify:** $$2\sqrt{2} + 2 - \sqrt{2} + 1 = (2\sqrt{2} - \sqrt{2}) + (2 + 1) = \sqrt{2} + 3$$ **Final answer:** $$\sqrt{2} + 3$$