Question

Simplifique as expressões em cada caso :
a)$( 3^{n} . 3^{n+1} . 3^{n+2}) : 3^{n+4}$[/tex]

b)$[ 2^{9} :( 2^{2} . 2 )^{2} ]^{-3}$

c) $[ 3^{4} . ( 3^{3} : 3^{2} )^{-1} ]^{-2}$

Answer 1

Olá, Clayton.

$a)~\frac{3^{n}\cdot3^{n+1}\cdot3^{n+2}}{3^{n+4}}=\frac{3^{n+n+1+n+2}}{3^{n+4}}=\frac{3^{3n+3}}{3^{n+4}}=3^{3n+3-n-4}=3^{2n-1}\\\\ b)~\left[\frac{2^{9}}{(2^{2}\cdot2)^{2}}\right]^{-3}= \left[\frac{2^{9}}{(2^3)^{2}}\right]^{-3}=\left[\frac{2^{9}}{2^6}\right]^{-3}=[2^{9-6}]^{-3}=[2^3]^{-3}=2^{3\cdot(-3)}=\\\\=2^{-9}=\frac1{2^9}=\frac1{512}$

$c)~[3^{4}\cdot( 3^{3} : 3^{2} )^{-1} ]^{-2}=[3^4\cdot(3^{3-2})^{-1}]^{-2}=[3^4\cdot(3^1)^{-1}]^{-2}=\\\\=[3^4\cdot3^{1\cdot(-1)}]^{-2}=[3^4\cdot3^{-1}]^{-2}=[3^{4-1}]^{-2}=[3^3]^{-2}=3^{3\cdot(-2)}=\\\\=3^{-6}=\frac1{3^6}=\frac1{729}$