# Question #fde5d

##### 1 Answer

#### Explanation:

We will use the property that

First, let's figure out what that value will be. Note that

#=5^2(y^2)^2*3xy#

#=(5y^2)^2*3xy#

Thus, to make it a perfect square, that is, to have all of the powers be even, we need to multiply by

So, using the property we mentioned initially, we can proceed:

#=sqrt(21x^2y*3xy)/sqrt(75xy^5*3xy)#

#=sqrt(63x^3y^2)/sqrt((15xy^3)^2)#

#=sqrt(63x^3y^2)/(15xy^3)#

The denominator is now rationalized, but we can do some further simplification by pulling out any squares from the remaining square root.

#=sqrt((3^2*7)(x^2*x)y^2)/(15xy^3)#

#=sqrt((3xy)^2*7x)/(15xy^3)#

#=(sqrt((3xy)^2)*sqrt(7x))/(15xy^3)#

#=(3xysqrt(7x))/(15xy^3)#

#=sqrt(7x)/(5y^2)#