Question

\(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}+\dfrac{2-\sqrt{2}}{\sqrt{2}-1}-\sqrt{18}\)

Answer 1

\(=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}+\dfrac{\sqrt{2}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}-3\sqrt{2}\)

\(=\sqrt{2}+\sqrt{2}-3\sqrt{2}=-\sqrt{2}\)

Answer 2

$\dfrac{2+\sqrt{2}}{\sqrt{2}+1}+\dfrac{2-\sqrt{2}}{\sqrt{2}-1}-\sqrt{18}\\=\dfrac{(2+\sqrt{2})(\sqrt{2}-1)}{(\sqrt{2}-1)(\sqrt{2}+1)}+\dfrac{(2-\sqrt{2})(\sqrt{2}+1)}{(\sqrt{2}-1)(\sqrt{2}+1)}-3\sqrt{2}\\=\dfrac{2\sqrt{2}+2-2-\sqrt{2}}{2-1}+\dfrac{2\sqrt{2}-2+2-\sqrt{2}}{2-1}-3\sqrt{2}\\=2\sqrt{2}-\sqrt{2}+2\sqrt{2}-\sqrt{2}-3\sqrt{2}\\=-\sqrt{2}$