George C. May 16, 2015 \displaystyle\sqrt{{\frac{{11}}{{25}}}}=\frac{\sqrt{{{11}}}}{\sqrt{{{25}}}}=\frac{\sqrt{{{11}}}}{{5}} In general \displaystyle\sqrt{{{a}{b}}}=\sqrt{{{a}}}\sqrt{{{b}}} ...

smendyka Mar 19, 2018 \displaystyle\sqrt{{\frac{{16}}{{36}}}}=\frac{\sqrt{{{16}}}}{\sqrt{{{36}}}}=\frac{{4}}{{6}}=\frac{{2}}{{3}}

\displaystyle\frac{{3}}{{4}} Explanation: First, simplify the fraction inside the square root by dividing by \displaystyle{2}{/}{2} . \displaystyle\sqrt{{\frac{{18}}{{32}}}}=\sqrt{{\frac{{9}}{{16}}}} ...

\displaystyle{\sqrt[{{3}}]{{\frac{{16}}{{27}}}}}=\frac{{{\sqrt[{{3}}]{{16}}}}}{{3}} Explanation: \displaystyle{16} is not a cube. so it does not have a rational cube root. However, \displaystyle{27} ...

\displaystyle\frac{{8}}{{5}} Explanation: We can change this problem to: \displaystyle\sqrt{{\frac{{64}}{{25}}}}\to\frac{\sqrt{{{64}}}}{\sqrt{{{25}}}}\to\frac{{8}}{{5}}

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