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Question #789d1

1 Answer

Jan 27, 2017

#= 1/(9x^-1)# or #x/9#

Explanation:

#((3x)^2(3x)^-3)/(3x^-2)#

#= (3x)^(2+(-3))/(3x^-2)#, use #a^m*a^n = a^(m+n)#

#= (3x)^-1/(3x^-2)=1/((3x)(3x^-2))#, use #a^-m=1/a^m#

#=1/(3*x*3*x^-2)#

#= 1/(9x^-1)# or #x/9#

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