smc

To make this problem easier to solve, lets get the radicals out of the denominator. For $\frac{1}{2+\sqrt{2}}$, we will multiply the numerator and denominator by $2-\sqrt{2}$ so, $\frac{1}{2+\sqrt{2}}\cdot \frac{2-\sqrt{2}}{2-\sqrt{2}}\Rightarrow \frac{2-\sqrt{2}}{2}$. Now, the other fraction we need to get the radical out of the denominator is $\frac{1}{\sqrt{2}-2}$. Here, we will multiply by the conjugate again, $\sqrt{2}+2$. So that simplifies to $\frac{1}{\sqrt{2}-2}\cdot \frac{\sqrt{2}+2}{\sqrt{2}+2}\Rightarrow \frac{\sqrt{2}+2}{-2}$. So now our simplified equation is $2+\sqrt{2}+\frac{2-\sqrt{2}}{2}+\frac{\sqrt{2}+2}{-2}\Rightarrow 2+\sqrt{2}+\frac{2-\sqrt{2}}{2}-\frac{\sqrt{2}+2}{2}$ Bringing everything to the same denominator and combining like terms, we get $\frac{4+2\sqrt{2}+2-\sqrt{2}-\sqrt{2}-2}{2}\Rightarrow \frac{4}{2}\Rightarrow 2\Rightarrow \boxed{2}$