Question

thực hiện phép tính

a) A=\(\dfrac{1}{\sqrt{2}\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

Answer 1

sữa đề chút nha :

ta có : \(A=\dfrac{1}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{1}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

\(\Leftrightarrow A=\dfrac{\sqrt{2}}{2+\sqrt{4+2\sqrt{3}}}+\dfrac{\sqrt{2}}{2-\sqrt{4-2\sqrt{3}}}\)

\(\Leftrightarrow A=\dfrac{\sqrt{2}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{\sqrt{2}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}\)

\(\Leftrightarrow A=\dfrac{\sqrt{2}}{3+\sqrt{3}}+\dfrac{\sqrt{2}}{3-\sqrt{3}}=\dfrac{\sqrt{2}\left(3-\sqrt{3}\right)+\sqrt{2}\left(3+\sqrt{3}\right)}{\left(3+\sqrt{3}\right)\left(3-\sqrt{3}\right)}\)

\(\Leftrightarrow A=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)