Question

Rút gọn: A=\(\frac{3+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\frac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)

Answer 1

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

Answer 2

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

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