How do you simplify #4/(sqrt(4x) + sqrt(4x))#?

1 Answer
Apr 22, 2018

#1/sqrtx#

Explanation:

#sqrt(4x)+sqrt(4x)=2sqrt4x#

Furthermore, recalling that #sqrt(ab)=sqrta*sqrtb,#

#2sqrt(4x)=2(sqrt4*sqrtx)#

#sqrt4=2#, so we get

#2(2sqrtx)=4sqrtx#

And we end up with

#cancel4/(cancel4sqrtx)=1/sqrtx#