An algebra problem by Mardokay Mosazghi

Adding the two fractions together first: $\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{\sqrt{2}-2}=\dfrac{2\sqrt{2}}{-2}=-\sqrt{2}$

Thus, $2+\sqrt{2}+\dfrac{1}{2+\sqrt{2}}+\dfrac{1}{\sqrt{2}-2}=2+\sqrt{2}-\sqrt{2}=\boxed{2}$

$\qquad \sqrt { 2 } +2+\frac { 1 }{ 2+\sqrt { 2 } } +\frac { 1 }{ \sqrt { 2 } -2 } \\ =\quad \quad \sqrt { 2 } +2+\frac { 1 }{ \sqrt { 2 } +2 } +\frac { 1 }{ \sqrt { 2 } -2 } \\ =\quad \quad \frac { -2(\sqrt { 2 } +2)+1(\sqrt { 2 } -2)+1(\sqrt { 2 } +2) }{ -2 } \\ \qquad \qquad \because (\sqrt { 2 } +2)(\sqrt { 2 } -2)=-2\\ =\quad \quad \frac { -2\sqrt { 2 } -4+\sqrt { 2 } -2+\sqrt { 2 } +2 }{ -2 } \\ =\quad \quad \frac { -2\sqrt { 2 } -4+2\sqrt { 2 } }{ -2 } \quad =\quad \frac { -4 }{ -2 } \quad =\quad \boxed { 2 }$