Resposta:

a) [tex]\frac{\sqrt{6}}{3}[/tex]

b) [tex]\frac{\sqrt{2}}{3}[/tex]

c) [tex]\frac{3 - \sqrt{3}}{6}[/tex]

d) [tex]\sqrt{10} + \sqrt{3}[/tex]

e) [tex]\frac{3+\sqrt{3}}{2}[/tex]

Explicação passo a passo:

a) [tex]\frac{2}{\sqrt{6}} = \frac{2}{\sqrt{6}} \times \frac{\sqrt{6}}{\sqrt{6}} = \frac{2\sqrt{6}}{6} = \frac{\sqrt{6}}{3}[/tex]

b) [tex]\frac{2}{3\sqrt{2}} = \frac{2}{3\sqrt2} \times \frac{\sqrt2}{\sqrt2} = \frac{2\sqrt2}{3\times2} = \frac{\sqrt2}{3}[/tex]

c) [tex]\frac{1}{3+\sqrt3} = \frac{1}{3+\sqrt3} \times \frac{(3-\sqrt3)}{(3-\sqrt3)} = \frac{3 - \sqrt3}{3^2 - (\sqrt3)^2} = \frac{3 - \sqrt3}{9-3} = \frac{3-\sqrt3}{6}[/tex]

d)

[tex]\frac{7}{\sqrt{10}-\sqrt3} = \frac{7}{\sqrt{10}-\sqrt3} \times \frac{(\sqrt{10}+\sqrt3)}{(\sqrt{10}+\sqrt3)} = \frac{7(\sqrt{10}+\sqrt3)}{\sqrt{10^2}-\sqrt{3^2}} = \\\\\frac{7(\sqrt{10}+\sqrt3)}{10-3} = \frac{7(\sqrt{10}+\sqrt3)}{7} = \sqrt{10} + \sqrt3[/tex]

e) [tex]\frac{\sqrt3}{\sqrt3-1} = \frac{\sqrt3}{\sqrt3-1} \times \frac{(\sqrt3+1)}{(\sqrt3+1)} = \frac{\sqrt3(\sqrt3+1)}{\sqrt{3^2} - 1^2} = \frac{\sqrt{3^2}+\sqrt3}{3-1} = \frac{3+\sqrt3}{2}[/tex]