Para racionalizar o denominador devemos deixar os valores com expoente 1:

a)

[tex]\dfrac{6}{\sqrt[4]{3} }= \dfrac{6}{\sqrt[4]{3} } \cdot\dfrac{\sqrt[4]{3^3} }{\sqrt[4]{3^3} } =\dfrac{6\sqrt[4]{3^3} }{\sqrt[4]{3^4} } =\dfrac{6\sqrt[4]{3^3} }{3 }[/tex]

b)

[tex]\dfrac{1}{\sqrt[5]{2^4} }=\dfrac{1}{\sqrt[5]{2^4} } \cdot\dfrac{\sqrt[5]{2} }{\sqrt[5]{2} } =\dfrac{\sqrt[5]{2} }{\sqrt[5]{2^5} } =\dfrac{\sqrt[5]{2} }{2 }[/tex]

c)

[tex]\dfrac{ab}{\sqrt[3]{a^2} } =\dfrac{ab}{\sqrt[3]{a^2} } \cdot\dfrac{\sqrt[3]{a} }{\sqrt[3]{a} } =\dfrac{ab\sqrt[3]{a} }{\sqrt[3]{a^3} } =\dfrac{b\sqrt[4]{a^3} }{a}[/tex]

d)

[tex]\dfrac{2x}{\sqrt[6]{x^5} } =\dfrac{2x}{\sqrt[6]{x^5} } \cdot\dfrac{\sqrt[6]{x} }{\sqrt[6]{x}} =\dfrac{2x\sqrt[6]{x}}{\sqrt[6]{x^6} } =\dfrac{2x\sqrt[6]{x}}{x} =2\sqrt[6]{x}[/tex]

e)

[tex]\dfrac{2}{\sqrt{7} +\sqrt{5} } =\dfrac{2}{\sqrt{7} +\sqrt{5} }\cdot\dfrac{\sqrt{7} -\sqrt{5} }{\sqrt{7} -\sqrt{5} } =\dfrac{2(\sqrt{7} -\sqrt{5} )}{7-5} =\dfrac{2(\sqrt{7} -\sqrt{5} )}{2} =\sqrt{7} -\sqrt{5}[/tex]

f)

[tex]\dfrac{7}{3-\sqrt{2} } =\dfrac{7}{3-\sqrt{2} } \cdot\dfrac{3+\sqrt{2} }{3+\sqrt{2}} =\dfrac{7(3+\sqrt{2})}{9-2} =\dfrac{7(3+\sqrt{2} )}{7} =3+\sqrt{2}[/tex]

g)

[tex]\dfrac{1}{\sqrt{7} -2} =\dfrac{1}{\sqrt{7} -2}\cdot\dfrac{\sqrt{7} +2 }{\sqrt{7} +2} =\dfrac{\sqrt{7} +2}{7-4} =\dfrac{\sqrt{7} +2}{3}[/tex]

h)

[tex]\dfrac{\sqrt{ab} }{\sqrt{a}-\sqrt{b}} =\dfrac{\sqrt{ab} }{\sqrt{a}-\sqrt{b}} \cdot\dfrac{\sqrt{a}+\sqrt{b}}{\sqrt{a}+\sqrt{b}} =\dfrac{\sqrt{ab}( \sqrt{a}+\sqrt{b})}{a-b}[/tex]

i)

[tex]\dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt2}} =\dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}+\sqrt2}} \cdot\dfrac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}} =\dfrac{(\sqrt{5}-\sqrt{2})^2}{5-2} =\dfrac{(\sqrt{5}-\sqrt{2})^2}{3} =[/tex]

j)

[tex]\dfrac{6+\sqrt{2}}{\sqrt{2}-1} =\dfrac{6+\sqrt{2}}{\sqrt{2}-1}\cdot\dfrac{\sqrt{2}+1}{\sqrt{2}+1} =\dfrac{(6+\sqrt{2})(\sqrt{2}+1)}{2-1} =8+7\sqrt{2}[/tex]

l)

[tex]\dfrac{\sqrt{a}+1}{\sqrt{a}-1} =\dfrac{\sqrt{a}+1}{\sqrt{a}-1} \cdot\dfrac{\sqrt{a}+1}{\sqrt{a}+1} =\dfrac{(\sqrt{a}+1)^2}{a-1}[/tex]

m)

[tex]\dfrac{5+\sqrt{11} }{3-\sqrt{11} } =\dfrac{5+\sqrt{11} }{3-\sqrt{11} } \cdot\dfrac{3+\sqrt{11} }{3+\sqrt{11}} =\dfrac{(5+\sqrt{11})(3+\sqrt{11})}{3-11} =-\dfrac{(4-2\sqrt{11} )}{8}[/tex]