Đặt: \(A=\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{\sqrt{2}}{2-\sqrt{4-2\sqrt{3}}}\)
\(=\dfrac{\sqrt{2}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{\sqrt{2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)
\(=\dfrac{\sqrt{2}}{3+\sqrt{3}}+\dfrac{\sqrt{2}}{3-\sqrt{3}}=\dfrac{3\sqrt{2}-\sqrt{6}+3\sqrt{2}+\sqrt{6}}{9-3}\)
\(=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)
\(\)
Đặt
\(A=\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}\);\(B=\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)
\(\Leftrightarrow A^2=\dfrac{1}{2+2+\sqrt{3}+2.\sqrt{2}.\sqrt{2+\sqrt{3}}}\)
\(\Leftrightarrow A^2=\dfrac{1}{4+\sqrt{3}+2\left(\sqrt{4+2\sqrt{3}}\right)}\)
\(\Leftrightarrow A^2=\dfrac{1}{4+\sqrt{3}+2\left(\sqrt{3}+1\right)}\)
\(\Leftrightarrow A^2=\dfrac{1}{6+3\sqrt{3}}\)
\(A=\dfrac{1}{\sqrt{6+3\sqrt{3}}}\)
Tương Tự Ta Có :
\(B=\dfrac{1}{\sqrt{6-3\sqrt{3}}}\)
\(\Rightarrow A+B=\dfrac{1}{\sqrt{6+3\sqrt{3}}}+\dfrac{1}{\sqrt{6-3\sqrt{3}}}\)
\(\Leftrightarrow\left(A+B\right)^2=\dfrac{1}{6+3\sqrt{3}}+\dfrac{1}{6-3\sqrt{3}}+\dfrac{2}{\sqrt{36-27}}\)
\(\Leftrightarrow\left(A+B\right)^2=\dfrac{6-3\sqrt{3}+6+3\sqrt{3}}{36-27}+\dfrac{2}{3}\)
\(\Leftrightarrow\left(A+B\right)^2=\dfrac{12}{9}+\dfrac{2}{3}=2\)
\(\Leftrightarrow A+B=\sqrt{2}\)
\(\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\sqrt{2}\)
