How do you simplify #sqrt(3x^5 )/ sqrt(8x²)#?

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Answer 1 · 2017-04-29T13:56:04

#(sqrt6)/4#

#(sqrt3x^5)/(sqrt8x^2)#

#:.=(sqrt(3*x*x*x*x*x ))/sqrt(2*2*2*x*x)#

#:.color(blue)(sqrtx xx sqrtx=x#

#:.color(blue)(sqrt2 xx sqrt2=2#

#:.=(x^2sqrt(3x))/(2x^2sqrt2)#

#:.=(sqrt(3x))/(2sqrt2)#

#:. color(blue)((2sqrt2)/(2sqrt2)=1/1 #

#:.=(sqrt(3x))/(2sqrt2) xx (2sqrt2)/(2sqrt2) #

#:.color(blue)(sqrt2 xx sqrt2=2#

#:.=(cancel2sqrt6)/cancel8^4#

#:.color(blue)(=sqrt6/4#

check:

Let #x=1#

#(sqrt(3*1^5))/(sqrt(8*1^2))#

#color(blue)((sqrt(3))/(sqrt(8))=0.612372435#

and #color(blue)(sqrt6/4=0.612372435#