a) Ta có: \(\frac{\sqrt{5}-2}{5+2\sqrt{5}}-\frac{1}{2+\sqrt{5}}+\frac{1}{\sqrt{5}}\)
\(=\frac{\sqrt{5}-2}{\sqrt{5}\left(\sqrt{5}+2\right)}-\frac{\sqrt{5}}{\sqrt{5}\left(\sqrt{5}+2\right)}+\frac{\sqrt{5}+2}{\sqrt{5}\left(\sqrt{5}+2\right)}\)
\(=\frac{\sqrt{5}-2-\sqrt{5}+\sqrt{5}+2}{\sqrt{5}\left(\sqrt{5}+2\right)}\)
\(=\frac{\sqrt{5}}{\sqrt{5}\left(\sqrt{5}+2\right)}\)
\(=\frac{1}{\sqrt{5}+2}\)
b) Ta có: \(\frac{1}{2+\sqrt{3}}+\frac{\sqrt{2}}{\sqrt{6}}-\frac{2}{3+\sqrt{3}}\)
\(=\frac{\sqrt{6}\left(\sqrt{3}+1\right)}{\sqrt{6}\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}+\frac{\sqrt{2}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}-\frac{2\sqrt{2}\cdot\left(\sqrt{3}+2\right)}{\sqrt{6}\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)
\(=\frac{3\sqrt{2}+\sqrt{6}+\sqrt{2}\cdot\left(5+3\sqrt{3}\right)-2\sqrt{6}-4\sqrt{2}}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)
\(=\frac{-\sqrt{2}-\sqrt{6}+5\sqrt{2}+3\sqrt{6}}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)
\(=\frac{4\sqrt{2}+2\sqrt{6}}{\sqrt{6}\cdot\left(\sqrt{3}+1\right)\left(\sqrt{3}+2\right)}\)
\(=\frac{2\sqrt{2}\cdot\left(2+\sqrt{3}\right)}{\sqrt{3}\cdot\sqrt{2}\cdot\left(2+\sqrt{3}\right)\left(\sqrt{3}+1\right)}\)
\(=\frac{2}{3+\sqrt{3}}\)
