Resposta :
a)[tex]\frac{\sqrt{15}}{\sqrt{3}} = \frac{\sqrt{15}}{\sqrt{3}} * \frac{\sqrt{3}}{\sqrt{3}} = \frac{\sqrt{15}*\sqrt{3}}{3} = \frac{\sqrt{45}}{3} = \frac{3\sqrt{5}}{3} = \sqrt{5}[/tex]
b)[tex]\frac{\sqrt{3}}{\sqrt{6}} = \frac{\sqrt{3}}{\sqrt{6}} * \frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{18}}{6} = \frac{3\sqrt{2}}{6} = \frac{\sqrt{2}}{2}[/tex]
c)[tex]\frac{\sqrt{18}}{\sqrt{6}} = \frac{\sqrt{18}}{\sqrt{6}}*\frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{108}}{6} = \frac{6\sqrt{3}}{6} = \sqrt{3}[/tex]
d)
e)
a) \frac{\sqrt{15}}{\sqrt3} = \frac{{\sqrt5}\cdot{\sqrt3}}{\sqrt3}[/tex]
simplificando temos[tex]\sqrt5[/tex]
b) [tex]\frac{\sqrt3}{\sqrt6} = \frac{\sqrt3}{{\sqrt2}\cdot{\sqrt3}}[/tex]
simplificando temos: [tex]\frac1{\sqrt3}[/tex]
c) [tex]\frac{\sqrt18}{\sqrt6} = \frac{{\sqrt6}\cdot{\sqrt3}}{\sqrt6}[/tex]
simplificando, temos: [tex]\sqrt3[/tex]
d) [tex]\sqrt[3]{2} \cdot \sqrt2 = \sqrt[6]{2}
e) \[tex]\frac{\sqrt[3]{4}}{\sqrt2} = \frac{\sqrt[3]{2^2}}{\sqrt2}[/tex]
[tex]\frac{2^\frac23}{2^\frac12} = 2^{\frac23 - \frac12}[/tex]
simplificando, temos: [tex]2^\frac16 = \sqrt[6]{2}[/tex]