Question

ayuda chicos please,si es posible con resolucion

Answer 1

Hola!

Respuesta:

8

Explicación paso a paso:

$\sqrt[b]{ \frac{ {6}^{a} . {16}^{b} . {3}^{a + 2b} }{ {18}^{a + b} } } \\ \\ = \sqrt[b]{ \frac{ {6}^{a}. {16}^{b} . {3}^{a} . {3}^{2b} } { {18}^{a}. {18}^{b} }} \\ \\ = \sqrt[b]{ \frac{ {(6.3)}^{a}. {16}^{b} . {3}^{2b} }{ {18}^{a}. {18}^{b} } } \\ \\ = \sqrt[b]{ \frac{ {18}^{a}. {16}^{b} . {3}^{2b} }{ {18}^{a}. {18}^{b} } } \\ \\ = \sqrt[b]{ \frac{ {16}^{b}. {3}^{b} . {3}^{b} }{ {18}^{b} } } \\ \\ = \sqrt[b]{ \frac{ {(16.3.3)}^{b} }{ {18}^{b} } } \\ \\ = \sqrt[b]{ \frac{ {144}^{b} }{ {18}^{b} } } \\ \\ = \sqrt[b]{ \frac{ {18}^{b}. {8}^{b} }{ {18}^{b} } } \\ \\ = \sqrt[b]{ {8}^{b} } \\ \\ = {8}^{ \frac{b}{b} } \\ \\ = 8$

Answer 2

Explicación paso a paso:

$\sqrt[b]{} \frac{3^{2a}.2^{a}.2^{4b}.3^{2b} }{}\frac{}{3^{2a}.2^{a}.3^{2b}.2^{b} }$=$\sqrt[b]{2^{3b} }$=$2^{3b/b}$=8