Ta có : \(\frac{1-\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}=\frac{\left(1-\sqrt{2}\right)\left(2\sqrt{3}+3\sqrt{2}\right)}{\left(2\sqrt{3}-3\sqrt{2}\right)\left(2\sqrt{3}+3\sqrt{2}\right)}=\frac{2\sqrt{3}+3\sqrt{2}-2\sqrt{6}-6}{12-18}\)

\(=\frac{\sqrt{12}+\sqrt{18}-\sqrt{24}-\sqrt{36}}{-6}\)\(=\frac{-\sqrt{12}-\sqrt{18}+\sqrt{24}+\sqrt{36}}{6}\)

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

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

\(=\frac{2\sqrt{3}+3\sqrt{2}-2\sqrt{6}-6}{12-18}\)

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

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

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

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

\(=\frac{2\sqrt{3}+3\sqrt{2}-2\sqrt{6}-6}{12-18}\)

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

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

Hông chắc !!!

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

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

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

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

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

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

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