Question

me ajudem pfvr urgentemente ​

Answer 1

Resposta:

4-

a)2$\sqrt{3}$

b)27

c)6$\sqrt[6]{48}$ ou 11,43821

5-

a)$\sqrt{2}$

b)$\sqrt[6]{2}$

c)4$\sqrt[3]{4}$

Explicação passo-a-passo:

4-

a)simplifica

2×6=12

$\sqrt{12}$

fatora em quadrado perfeito

$\sqrt{2^{2} } .3$

$\sqrt{2^{2} } \sqrt{3}$

b)simplifica tudo e depois calcula

$\sqrt{3}$×3×3$\sqrt{3}$

3×3×3

27

c)expande as expressoes

$\sqrt[6]{12^{3} } \sqrt[6]{36^{2} }$

$\sqrt[6]{12^{3}.36^{2} }$

simplifica

$\sqrt[6]{1728.36^{2} }$

$2\sqrt[6]{27.36^{2} }$

multiplica

$2\sqrt[6]{27.1296}$

simplifica tudo

$2\sqrt[6]{34992}$

6$\sqrt[6]{48}$

5-

a)simplifica

$\frac{\sqrt[6]{16} \sqrt[6]{8} }{\sqrt[6]{18} }$

corta os iguais

$\sqrt[6]{8}$

simplifica

$\sqrt[6]{2^{3} }$

$\sqrt{2}$

b)multiplica por $\frac{\sqrt{2} }{\sqrt{2} }$

multiplica as duas fracoes

$\frac{\sqrt[3]{4} }{\sqrt{2} } .\frac{\sqrt{2} }{\sqrt{2} }$

$\frac{ \sqrt[6]{4^{2} \sqrt[6]{2^{3} } } }{2}$

$\frac{\sqrt[6]{2^{4}.2^{3} } }{2}$

simplifica

$\frac{2\sqrt[6]{2^{7} } }{2}$

reduz

$\frac{2\sqrt[6]{2^{} } }{2}$

$\sqrt[6]{2}$

c)simplifica

$\frac{16}{2\sqrt[3]{2} }$

reduz

$\frac{8}{\sqrt[3]{2} }$

multiplica as duas fracoes

$\frac{8}{\sqrt[3]{2} } . \frac{\sqrt[3]{2^{2} } }{\sqrt[3]{2^{2} } }$

calcula

$\frac{8\sqrt[3]{2^{2} } }{\sqrt[3]{2.2^{2} } }$

simplifica

$\frac{8\sqrt[3]{2^{2} } }{\sqrt[3]{2^{3} } }$

reduz

$\frac{8\sqrt[3]{2^{2} } }{2}$

avalie

$4\sqrt[3]{2^{2} }$

4$\sqrt[3]{4}$

deem 5 estrelas pq demorou :)

espero ter ajudade;)