Answer
$\dfrac{7x^{2}}{5y^{3}}$
Work Step by Step
$\bf{\text{Solution Outline:}}$
To simplify the given expression, $
\left( \dfrac{49x^5y^3}{25xy^9} \right)^{\frac{1}{2}}
,$ use the laws of exponents.
$\bf{\text{Solution Details:}}$
Using the Quotient Rule of the laws of exponents, which states that $\dfrac{x^m}{x^n}=x^{m-n},$ the expression above simplifies to
\begin{array}{l}\require{cancel}
\left( \dfrac{49x^5y^3}{25xy^9} \right)^{\frac{1}{2}}
\\\\
\left( \dfrac{49x^{5-1}y^{3-9}}{25} \right)^{\frac{1}{2}}
\\\\
\left( \dfrac{49x^{4}y^{-6}}{25} \right)^{\frac{1}{2}}
.\end{array}
Using the Power of a Quotient Rule of the laws of exponents, which is given by $\left( \dfrac{x^m}{y^n} \right)^p=\dfrac{x^{mp}}{y^{np}},$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\left( \dfrac{7^2x^{4}y^{-6}}{5^2} \right)^{\frac{1}{2}}
\\\\=
\dfrac{7^{2\cdot\frac{1}{2}}x^{4\cdot\frac{1}{2}}y^{-6\cdot\frac{1}{2}}}{5^{2\cdot\frac{1}{2}}}
\\\\=
\dfrac{7^{1}x^{2}y^{-3}}{5^{1}}
\\\\=
\dfrac{7x^{2}y^{-3}}{5}
.\end{array}
Using the Negative Exponent Rule of the laws of exponents, which states that $x^{-m}=\dfrac{1}{x^m}$ or $\dfrac{1}{x^{-m}}=x^m,$ the expression above is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{7x^{2}y^{-3}}{5}
\\\\=
\dfrac{7x^{2}}{5y^{3}}
.\end{array}
