Question #fde5d

We will use the property that #sqrt(a)sqrt(b) = sqrt(ab)# to eliminate the radical from the denominator. To do so, we will multiply the numerator and denominator by such a value as to make the denominator the square root of a perfect square. With that, we will apply #sqrt(a^2) = a# for #a>0#.

First, let's figure out what that value will be. Note that #(x^a)^b = x^(ab)#.

#75xy^5 = 3(5^2)xy^5#

#=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 #3xy#:

#(5y^2)^2*3xy*3xy = (5y^2)^2*(3xy)^2 = (15xy^3)^2#

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

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

#=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)#