Đặt \(A=\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{2+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(\dfrac{1}{\sqrt{2}}A=\dfrac{2+\sqrt{5}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{2-\sqrt{5}}{2-\sqrt{6-2\sqrt{5}}}\)
\(\dfrac{1}{\sqrt{2}}A=\dfrac{2+\sqrt{5}}{2+\sqrt{5}+1}+\dfrac{2-\sqrt{5}}{2-\sqrt{5}+1}\)
\(\dfrac{1}{\sqrt{2}}A=\dfrac{2+\sqrt{5}}{3+\sqrt{5}}+\dfrac{2-\sqrt{5}}{3-\sqrt{5}}\)
(đề có sai không bạn nhỉ?)
Mình là bạn :vvv nhé;-;
Mình nghĩ nên sửa đề thành như này.
\(A=\dfrac{3-\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{3-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(\dfrac{1}{\sqrt{2}}A=\dfrac{3+\sqrt{5}}{2+\sqrt{6+2\sqrt{5}}}+\dfrac{3-\sqrt{5}}{2-\sqrt{6-2\sqrt{5}}}\)
\(\dfrac{1}{\sqrt{2}}A=\dfrac{3+\sqrt{5}}{2+\sqrt[]{5}+1}+\dfrac{3-\sqrt{5}}{2-\sqrt{5}+1}\)
\(\dfrac{1}{\sqrt{2}}A=1+1\)
\(A=2\sqrt{2}\)
Ta có: \(\dfrac{2+\sqrt{5}}{\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{2-\sqrt{5}}{\sqrt{2}-\sqrt{3-\sqrt{5}}}\)
\(=\dfrac{2\sqrt{2}+\sqrt{10}}{3+\sqrt{5}}+\dfrac{2\sqrt{2}-\sqrt{10}}{3-\sqrt{5}}\)
\(=\dfrac{\left(2\sqrt{2}+\sqrt{10}\right)\left(3-\sqrt{5}\right)+\left(2\sqrt{2}-\sqrt{10}\right)\left(3+\sqrt{5}\right)}{4}\)
\(=\dfrac{6\sqrt{2}-2\sqrt{10}+3\sqrt{10}-5\sqrt{2}+6\sqrt{2}+2\sqrt{10}-3\sqrt{10}-5\sqrt{2}}{4}\)
\(=\dfrac{2\sqrt{2}}{4}=\dfrac{\sqrt{2}}{2}\)