Question

\(\dfrac{6\sqrt{2}-4}{\sqrt{2}-3}\)

Answer 1

\(=\dfrac{\sqrt{72}-\sqrt{16}}{\sqrt{2}-3}=\dfrac{-\sqrt{8}\left(\sqrt{2}-\sqrt{9}\right)}{\sqrt{2}-3}=\dfrac{-2\sqrt{2}\left(\sqrt{2}-3\right)}{\sqrt{2}-3}=-2\sqrt{2}\)

Answer 2

\(\dfrac{6\sqrt{2}-4}{\sqrt{2}-3}=-2\sqrt{2}\)