Question

\(\dfrac{2}{3+2\sqrt{2}}-\dfrac{2}{3-\sqrt{8}}\)

Answer 1

\(=6-4\sqrt{2}-6-4\sqrt{2}=-8\sqrt{2}\)

Answer 2

\(\dfrac{2}{3+2\sqrt{2}}-\dfrac{2}{3-\sqrt{8}}\\ =\dfrac{2}{3+2\sqrt{2}}-\dfrac{2}{3-2\sqrt{2}}\\ =\dfrac{2\left(3-2\sqrt{2}\right)-2\left(3+2\sqrt{2}\right)}{\left(3-2\sqrt{2}\right)\left(3+2\sqrt{2}\right)}\\ =\dfrac{6-4\sqrt{2}-6-4\sqrt{2}}{3^2-\left(2\sqrt{2}\right)^2}\\ =\dfrac{-8\sqrt{2}}{9-8}\\ =\dfrac{-8\sqrt{2}}{1}\\ =-8\sqrt{2}\)