## Elementary and Intermediate Algebra: Concepts & Applications (6th Edition)

$\dfrac{\sqrt{6xy}}{3y}$
Multiplying by an expression equal to $1$ which will make the denominator a perfect power of the index results to \begin{array}{l}\require{cancel} \sqrt{\dfrac{2x}{3y}} \\\\= \sqrt{\dfrac{2x}{3y}\cdot\dfrac{3y}{3y}} \\\\= \sqrt{\dfrac{1}{(3y)^2}\cdot6xy} \\\\= \dfrac{1}{3y}\sqrt{6xy} \\\\= \dfrac{\sqrt{6xy}}{3y} .\end{array}