Chapter 9 - Roots and Radicals - 9.3 - More on Simplifying Radicals - Problem Set 9.3 - Page 412: 29

$\dfrac{2\sqrt{3}}{9}$

Recall, we are not allowed to have radicals in the denominator of a fraction. Using the properties of radicals, the given expression, $ \dfrac{\sqrt{4}}{\sqrt{27}} ,$ simplifies to \begin{array}{l}\require{cancel} \dfrac{\sqrt{4}}{\sqrt{27}}\cdot\dfrac{\sqrt{3}}{\sqrt{3}} \\\\= \dfrac{\sqrt{4\cdot3}}{\sqrt{81}} \\\\= \dfrac{\sqrt{(2)^2\cdot3}}{\sqrt{(9)^2}} \\\\= \dfrac{2\sqrt{3}}{9} .\end{array}