How do you simplify #25/sqrt125#?

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Answer 1 · 2017-05-30T17:12:15

#5/(sqrt(5)#

#25/sqrt(125)#

Let's rewrite the denominator.

#25/((sqrt(25))(sqrt(5)))#

Simplify

#25/(5(sqrt(5)))#

#5/(sqrt(5)#

Rationalise the denominator

#= frac(5)(sqrt(5)) cdot frac(sqrt(5))(sqrt(5))#

#= frac(5 sqrt(5))(5)#

#= sqrt(5)#

Answer 2 · 2017-05-30T17:26:25

#sqrt(5)#

We have: #frac(25)(sqrt(125))#

Let's express the denominator as a product:

#= frac(25)(sqrt(5^(2) cdot 5))#

#= frac(25)(5 sqrt(5))#

#= frac(5)(sqrt(5))#

Then, let's rationalise the denominator:

#= frac(5)(sqrt(5)) cdot frac(sqrt(5))(sqrt(5))#

#= frac(5 sqrt(5))(5)#

#= sqrt(5)#