Question

Simplify, step-by-step, sqrt (3x/2y^3)

Fan + Medal

Answer 1

\(\huge \sqrt{\frac{3x}{2y^3}} = \sqrt{\frac{3x}{2*y*y^2}} =
\frac 1y\sqrt{\frac{3x}{2y}}\)

Answer 2

the answer is \[1/2y ^{2} \sqrt{6xy}\], idk how they got it

Answer 3

Your answer and mine are the same but some books do not like the radical sign in the denominator and try to get rid of it. It is called "rationalizing the denominator" which I will do below:

Answer 4

\(\huge \sqrt{\frac{3x}{2y^3}} = \sqrt{\frac{3x}{2*y*y^2}} =
\frac 1y\sqrt{\frac{3x}{2y}} = \\ \huge
\frac 1y\sqrt{\frac{3x*2y}{2y*2y}} = \frac 1y\sqrt{\frac{6xy}{(2y)^2}} = \\ \huge
\frac 1y * \frac{1}{2y} * \sqrt{6xy} = \frac {1}{2y^2}\sqrt{6xy}
\)

Answer 5

ok how would I start this, \[\sqrt[5]{5x ^{7}y ^{2}/27x ^{2}}\]

Answer 6

@aum

Answer 7

instead of 7 for the exponent of 5x its 8

Answer 8

Please open up a new post for the second question.