Đặt A=\(\frac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(\Leftrightarrow\frac{1}{\sqrt{2}}A=\frac{1}{\sqrt{2}\left(\sqrt{2}+\sqrt{2+\sqrt{3}}\right)}+\frac{1}{\sqrt{2}\left(\sqrt{2}-\sqrt{2-\sqrt{3}}\right)}\)
\(=\frac{1}{2+\sqrt{4+2\sqrt{3}}}+\frac{1}{2-\sqrt{4-2\sqrt{3}}}\)
=\(\frac{1}{2+\sqrt{3+2\sqrt{3}+1}}+\frac{1}{2-\sqrt{3-2\sqrt{3}+1}}\)
\(=\frac{1}{2+\sqrt{3}+1}+\frac{1}{2-\sqrt{3}+1}\)
\(=\frac{1}{3+\sqrt{3}}+\frac{1}{3-\sqrt{3}}\)
\(=\frac{3-\sqrt{3}+3+\sqrt{3}}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)
\(=\frac{6}{9-3}\)
= 1
\(\Rightarrow A=\sqrt{2}\)
