Question

Tính:

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

Answer 1

Lời giải:

Ta có: \(\sqrt{2+\sqrt{3}}=\frac{\sqrt{4+2\sqrt{3}}}{\sqrt{2}}=\frac{\sqrt{3+1+2\sqrt{1.3}}}{\sqrt{2}}=\frac{\sqrt{(\sqrt{3}+1)^2}}{\sqrt{2}}=\frac{\sqrt{3}+1}{\sqrt{2}}\)

Do đó:

Gọi biểu thức đã cho là $A$ thì:

\(A=\frac{\sqrt{3}+1}{2\sqrt{2}}\left(\frac{\sqrt{3}+1}{2\sqrt{2}}-\frac{2}{\sqrt{6}}+\frac{\sqrt{3}+1}{2\sqrt{2}}\right)\)

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

\(=\frac{\sqrt{3}+1}{2\sqrt{2}}.\frac{3\sqrt{3}+3-\sqrt{12}}{3\sqrt{2}}=\frac{\sqrt{3}+1}{2\sqrt{2}}.\frac{3+\sqrt{3}}{3\sqrt{2}}\)

\(=\frac{(\sqrt{3}+1)^2.\sqrt{3}}{12}=\frac{(\sqrt{3}+1)^2}{4\sqrt{3}}\)