Question

Rút gọn

\(\frac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\frac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}\)

Answer 1

- Biến đổi \(2+\sqrt{3}=\frac{4+2\sqrt{3}}{2}=\frac{\left(\sqrt{3}+1\right)^2}{2}\)

- Tương tự \(2-\sqrt{3}=\frac{\left(\sqrt{3}-1\right)^2}{2}\)

Vậy A \(=\frac{\left(\sqrt{3}+1\right)^2}{2\sqrt{2}+\sqrt{6}+\sqrt{2}}+\frac{\left(\sqrt{3}-1\right)^2}{2\sqrt{2}-\sqrt{6}+\sqrt{2}}\)

\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{6}}=\sqrt{2}\)