## Algebra 2 Common Core

$\dfrac{2\sqrt{a}}{3ab}$
The variables are assumed to be non-zero as they are also in the denominator. RECALL: $\dfrac{\sqrt{a}}{\sqrt{b}} = \sqrt{\dfrac{a}{b}}, a\ge0, b\gt0.$ Use the rule above to have $\require{cancel} \sqrt{\dfrac{20ab}{45a^2b^3}} \\=\sqrt{\dfrac{\cancel{20}4a\cancel{b}}{\cancel{45}9a^2\cancel{b^3}b^2}} \\=\sqrt{\dfrac{4a}{9a^2b^2}} \\=\sqrt{\dfrac{2^2(a)}{(3ab)^2}} \\=\dfrac{2}{3ab} \cdot \sqrt{a} \\=\dfrac{2\sqrt{a}}{3ab}.$