Question

1. Tính A= \(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

Answer 1

\(A=\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{\sqrt{2}\left(2-\sqrt{3}\right)}{2-\sqrt{4-2\sqrt{3}}}\)

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

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

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

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