Question #d5bad

1 Answer
Nityananda
Oct 15, 2017

#x/(2y^4)#

Explanation:

Given, #(4x^-2. y^8)^(-1/2)#

#rArr 4^(-1/2). (x^-2)^(-1/2). (y^8)^(-1/2)#

#rArr (2^2)^(-1/2). x^[(-2)(-1/2)]. y^[8(-1/2)]# [As #(a^m)^n = a^(mn)]#

#rArr 2^-1. x^1. y^(-4)# [ As #a^-1 = 1/a]#

#rArr x/(2y^4)#

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