Question

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

Answer 1

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

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

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

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

\(=\sqrt{\dfrac{2}{3}}\cdot\dfrac{2\sqrt{3}}{2}=\sqrt{2}\)