Question

\(K=\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}\)

Answer 1

Ta có K=\(\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}\) =\(\dfrac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{4+2\sqrt{3}}}\) =\(\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}\) =\(\dfrac{\sqrt{2}\left(2+\sqrt{3}\right)}{3+\sqrt{3}}\)

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