Intermediate Algebra (6th Edition)

Published by Pearson
ISBN 10: 0321785045
ISBN 13: 978-0-32178-504-6

Chapter 7 - Section 7.5 - Rationalizing Denominators and Numerators of Radical Expressions - Exercise Set - Page 445: 30

Answer

$\dfrac{\sqrt[]{3}}{3}$

Work Step by Step

Multiplying by an expression equal to $1$ such that the denominator becomes a perfect power of the index, the given expression, $ \sqrt[4]{\dfrac{1}{9}} ,$ is equivalent to \begin{array}{l}\require{cancel} \sqrt[4]{\dfrac{1}{9}\cdot\dfrac{9}{9}} \\= \sqrt[4]{\dfrac{9}{81}} \\= \sqrt[4]{\dfrac{9}{3^4}} \\= \sqrt[4]{\dfrac{1}{3^4}\cdot9} \\= \sqrt[4]{\left(\dfrac{1}{3}\right)^4\cdot9} \\= \dfrac{1}{3}\sqrt[4]{9} \\= \dfrac{1}{3}\sqrt[4]{3^2} \\= \dfrac{1}{3}\sqrt[\frac{4}{2}]{3^{\frac{2}{2}}} \\= \dfrac{1}{3}\sqrt[2]{3^{1}} \\= \dfrac{1}{3}\sqrt[]{3} \\= \dfrac{\sqrt[]{3}}{3} .\end{array}
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