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