Question

\(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{3}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)

Answer 1

\(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{3}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\\ =\left(2-\sqrt{3}\right)+\dfrac{\sqrt{2}}{2}-\dfrac{2\left(3-\sqrt{3}\right)}{6}=\dfrac{12-6\sqrt{3}+3\sqrt{2}-6+2\sqrt{3}}{6}=\dfrac{6-\sqrt{3}}{6}\)

Answer 2

Ta có: \(\dfrac{1}{2+\sqrt{3}}+\dfrac{\sqrt{3}}{\sqrt{6}}-\dfrac{2}{3+\sqrt{3}}\)

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

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

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

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

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