Question

[Nguyễn Thị Thu Hằng]

Tính :

a, \(B=\sqrt{3-\sqrt{5}}\left(\sqrt{10}-\sqrt{2}\right)\left(3+\sqrt{5}\right)\)

b, \(A=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

c, \(C=\frac{1}{\sqrt{3}}+\frac{1}{3\sqrt{2}}-\frac{\sqrt{5-2\sqrt{6}}}{3}\)

Answer 1

[Nguyễn Việt Lâm]

\(B=\sqrt{6-2\sqrt{5}}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

\(=\sqrt{\left(\sqrt{5}-1\right)^2}\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)\)

\(=\left(\sqrt{5}-1\right)^2\left(3+\sqrt{5}\right)=\left(6-2\sqrt{5}\right)\left(3+\sqrt{5}\right)\)

\(=2\left(3-\sqrt{5}\right)\left(3+\sqrt{5}\right)=8\)

\(A^2=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}=8+2\sqrt{6-2\sqrt{5}}\)

\(A^2=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)

\(A^2=\left(\sqrt{5}+1\right)^2\Rightarrow A=\sqrt{5}+1\) (do \(A>0\))

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

\(=\frac{\sqrt{2}}{6}+\frac{\sqrt{2}}{3}=\frac{\sqrt{2}}{2}\)