\(\frac{1}{\sqrt{3}-1}-\frac{1}{\sqrt{3}+1}\)

\(=\frac{\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}-\frac{\sqrt{3}-1}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}\)

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

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

\(=\frac{2}{3-1}=\frac{2}{2}=1\)

Quy đồng lên ta có:
\(\frac{\sqrt{3}+1-\sqrt{3}+1}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)}\)

Áp dụng hằng đẳng thức ta có

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

\(\frac{1}{1+\sqrt{2}+\sqrt{3}}\)

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

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

\(=\frac{2+\sqrt{2}-\sqrt{6}}{4}\)

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

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

Ta có: \(\frac{1}{\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}}=\)\(\frac{\sqrt[3]{3}+\sqrt[3]{2}}{\left(\sqrt[3]{2}+\sqrt[3]{3}\right)\left(\sqrt[3]{9}-\sqrt[3]{6}+\sqrt[3]{4}\right)}=\frac{\sqrt[3]{2}+\sqrt[3]{3}}{\left(\sqrt[3]{2}\right)^3+\left(\sqrt[3]{3}\right)^3}=\frac{\sqrt[3]{2}+\sqrt[3]{3}}{5}\)

\(\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\left(\sqrt{2}+\sqrt{3}+\sqrt{5}\right)\left(\sqrt{2}+\sqrt{3}-\sqrt{5}\right)}\)
\(=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\left(\sqrt{2}+\sqrt{3}\right)^2-\left(\sqrt{5}\right)^2}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2+2\sqrt{2\cdot3}+3-5}=\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{5+2\sqrt{6}-5}=\frac{\sqrt{6}\cdot\left(\sqrt{2}+\sqrt{3}-\sqrt{5}\right)}{\sqrt{6}\cdot2\sqrt{6}}=\frac{2\sqrt{3}+3\sqrt{2}-\sqrt{30}}{12}\)

Ta có \(\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}\) = \(\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{5+2\sqrt{6}-5}\)

= \(\frac{\sqrt{6}\left(\sqrt{2}+\sqrt{3}-\sqrt{5}\right)}{12}\)