Chapter 8 - Section 8.2 - Multiplying and Dividing Radicals - Exercise Set - Page 585: 13

$\sqrt{x}$

Using $\sqrt{x}\cdot\sqrt{y}=\sqrt{xy}$ or the product rule of radicals, the given expression, $ \sqrt{\dfrac{2x}{9}}\cdot\sqrt{\dfrac{9}{2}} ,$ is equivalent to \begin{array}{l}\require{cancel} \sqrt{\dfrac{2x}{9}\cdot\dfrac{9}{2}} \\\\= \sqrt{\dfrac{\cancel{2}x}{\cancel{9}}\cdot\dfrac{\cancel{9}}{\cancel{2}}} \\\\= \sqrt{x} .\end{array}