Question

Giúp mk với!!!

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

Answer 1

Ta có: \(C=\dfrac{\sqrt{2+\sqrt{3}}}{2}:\left(\dfrac{\sqrt{2+\sqrt{3}}}{2}-\dfrac{2}{\sqrt{6}}+\dfrac{\sqrt{2+\sqrt{3}}}{2\sqrt{3}}\right)\)

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

\(=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\left(\dfrac{\sqrt{12+6\sqrt{3}}-4+\sqrt{4+2\sqrt{3}}}{2\sqrt{6}}\right)\)

\(=\dfrac{\sqrt{3}+1}{2\sqrt{2}}:\dfrac{3+\sqrt{3}-4+\sqrt{3}+1}{2\sqrt{6}}\)

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

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

Answer 2

ok