Respostas

2013-06-03T21:35:55-03:00

a)\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}

 

b)\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}

 

c)\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}

 

d)

 

e)

1 5 1
2013-06-03T22:07:02-03:00

 a)    \frac{\sqrt{15}}{\sqrt3} = \frac{{\sqrt5}\cdot{\sqrt3}}{\sqrt3}[/tex]

simplificando temos\sqrt5

 

b)   \frac{\sqrt3}{\sqrt6} = \frac{\sqrt3}{{\sqrt2}\cdot{\sqrt3}}

simplificando temos: \frac1{\sqrt3}

 

c)    \frac{\sqrt18}{\sqrt6} = \frac{{\sqrt6}\cdot{\sqrt3}}{\sqrt6}

simplificando, temos: \sqrt3

 

d)   \sqrt[3]{2} \cdot \sqrt2 = \sqrt[6]{2}</p>&#10;<p> </p>&#10;<p>e)   \[tex]\frac{\sqrt[3]{4}}{\sqrt2} = \frac{\sqrt[3]{2^2}}{\sqrt2}

       \frac{2^\frac23}{2^\frac12} = 2^{\frac23 - \frac12}

simplificando, temos: 2^\frac16 = \sqrt[6]{2}