How do I simplify #((x^2)/(4y^-2))^(-1/2)#?

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Answer 1 · 2018-07-26T16:14:31

#2/{xy}#

#(x^2/{4y^{-2}})^{-1/2}#

#=({x^2y^2}/{4})^{-1/2}#

#=({(xy)^2}/{4})^{-1/2}#

#={((xy)^2)^{-1/2}}/{4^{-1/2}}#

#=4^{1/2}(xy)^{2(-1/2)}#

#=(2^2)^{1/2}(xy)^{-1}#

#=\frac{2^{2(1/2)}}{xy}#

#=2/{xy}#

Answer 2 · 2018-07-26T16:29:48

#(x^2/(4y^(-2)))^(-1/2)=2/(xy)#

Let ,

#A=(x^2/(4y^(-2)))^(-1/2)#

#:.A=(x^2)^(-1/2)/((4y^(-2))^(-1/2)).....to[because(a/b)^n=a^n/b^n]#

#:.A#=#x^(2xx(-1/2))/(4^(-1/2)*y^((-2)(-1/2)))to[because(a^m)^n=a^(mn),(ab)^n=a^n*b^n]#

#:.A=x^(-1)/((2^2)^(-1/2)y^1#

#:.A=(1/x)/(2^-1 y)#

#:.A=2^1 xx 1/(xy)#

#:.A=2/(xy)#