Algebra

$\dfrac{1}{\sqrt2+\sqrt 3}+\dfrac{1}{\sqrt 2-\sqrt 3}\\ =\dfrac{\sqrt 2-\sqrt 3+\sqrt 2+\sqrt 3}{2-3}\\ =\dfrac{2\sqrt{2}}{-1}\\ \color{blue}=-2\sqrt{2}\ or =-\sqrt{8}$

  !