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\(\frac{1}{1-\sqrt[3]{2}}=\frac{\left(1+\sqrt[3]{2}+\sqrt[3]{4}\right)}{\left(1-\sqrt[3]{2}\right)\left(1+\sqrt[3]{2}+\sqrt[3]{4}\right)}=\frac{1+\sqrt[3]{2}+\sqrt[3]{4}}{-1}\)
\(=-1-\sqrt[3]{2}-\sqrt[3]{4}\)
\(=\frac{\left(\sqrt[3]{2^2}+\sqrt[3]{2}+1\right)}{\left(1-\sqrt[3]{2}\right)\left(\left(\sqrt[3]{2^2}+\sqrt[3]{2}+1\right)\right)}\)
=\(\frac{\left(\sqrt[3]{2^2}+\sqrt[3]{2}+1\right)}{1-2}\)
\(-\left(\sqrt[3]{2^2}+\sqrt[3]{2}+1\right)\)
\(\frac{1}{1+\sqrt{2}+\sqrt{3}}\)
\(=\frac{1+\sqrt{2}-\sqrt{3}}{\left(1+\sqrt{2}+\sqrt{3}\right)\left(1+\sqrt{2}-\sqrt{3}\right)}\)
\(=\frac{1+\sqrt{2}-\sqrt{3}}{2\sqrt{2}}\)
\(=\frac{2+\sqrt{2}-\sqrt{6}}{4}\)
\(\frac{1}{\sqrt{2}+\sqrt{3}}\\ =\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}+\sqrt{2}\right)\left(\sqrt{3}-\sqrt{2}\right)}\\ =\frac{\sqrt{3}-\sqrt{2}}{\left(\sqrt{3}\right)^2-\left(\sqrt{2}\right)^2}\\ =\frac{\sqrt{3}-\sqrt{2}}{1}=\sqrt{3}-\sqrt{2}\)
\(\frac{1}{\sqrt{2}+\sqrt{3}}=\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}+\sqrt{3}\right)\left(\sqrt{2}-\sqrt{3}\right)}=\frac{\sqrt{2}-\sqrt{3}}{\left(\sqrt{2}\right)^2-\left(\sqrt{3}\right)^2}=\frac{\sqrt{2}-\sqrt{3}}{2-3}=\frac{\sqrt{2}-\sqrt{3}}{-1}=-\left(\sqrt{2}-\sqrt{3}\right)=-\sqrt{2}+\sqrt{3}\)
\(\dfrac{1}{3\sqrt{2}+3\sqrt{4}+1}=\dfrac{1}{7+3\sqrt{2}}=\dfrac{7-3\sqrt{2}}{49-18}=\dfrac{7-3\sqrt{2}}{31}\)
\(\dfrac{1}{3\sqrt{2}+3\sqrt{4}+1}=\dfrac{1}{3\sqrt{2}+3.2+1}=\dfrac{1}{3\sqrt{2}+7}=\dfrac{3\sqrt{2}-7}{\left(3\sqrt{2}+7\right)\left(3\sqrt{2}-7\right)}=\dfrac{3\sqrt{2}-7}{18-49}=\dfrac{7-3\sqrt{2}}{31}\)
Ta có : \(\frac{1-\sqrt{2}}{2\sqrt{3}-3\sqrt{2}}=\frac{\left(1-\sqrt{2}\right)\left(2\sqrt{3}+3\sqrt{2}\right)}{\left(2\sqrt{3}-3\sqrt{2}\right)\left(2\sqrt{3}+3\sqrt{2}\right)}=\frac{2\sqrt{3}+3\sqrt{2}-2\sqrt{6}-6}{12-18}\)
\(=\frac{\sqrt{12}+\sqrt{18}-\sqrt{24}-\sqrt{36}}{-6}\)\(=\frac{-\sqrt{12}-\sqrt{18}+\sqrt{24}+\sqrt{36}}{6}\)
