Question #d5bad

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Question #d5bad

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Answer 1 · 2017-10-15T14:31:28

$\frac{x}{2 y^{4}}$ $x/(2y^4)$

Explanation:

Given, ${(4 x^{- 2} . y^{8})}^{- \frac{1}{2}}$ $(4x^-2. y^8)^(-1/2)$

$\Rightarrow 4^{- \frac{1}{2}} . {(x^{- 2})}^{- \frac{1}{2}} . {(y^{8})}^{- \frac{1}{2}}$ $rArr 4^(-1/2). (x^-2)^(-1/2). (y^8)^(-1/2)$

$\Rightarrow {(2^{2})}^{- \frac{1}{2}} . x^{(- 2) (- \frac{1}{2})} . y^{8 (- \frac{1}{2})}$ $rArr (2^2)^(-1/2). x^[(-2)(-1/2)]. y^[8(-1/2)]$ [As ${(a^{m})}^{n} = a^{m n}]$ $(a^m)^n = a^(mn)]$

$\Rightarrow 2^{- 1} . x^{1} . y^{- 4}$ $rArr 2^-1. x^1. y^(-4)$ [ As $a^{- 1} = \frac{1}{a}]$ $a^-1 = 1/a]$

$\Rightarrow \frac{x}{2 y^{4}}$ $rArr x/(2y^4)$

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