Answer
$4xy^{4}-2y+\dfrac{1}{x^{2}y}$
Work Step by Step
Dividing each of the terms of the numerator by the denominator, the given expression, $
\dfrac{16x^3y^5-8x^2y^2+4}{4x^2y}
,$ is equivalent to
\begin{array}{l}\require{cancel}
\dfrac{16x^3y^5}{4x^2y}-\dfrac{8x^2y^2}{4x^2y}+\dfrac{4}{4x^2y}
.\end{array}
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}
4x^{3-2}y^{5-1}-2x^{2-2}y^{2-1}+x^{-2}y^{-1}
\\\\=
4x^{1}y^{4}-2x^{0}y^{1}+x^{-2}y^{-1}
\\\\=
4xy^{4}-2y+x^{-2}y^{-1}
.\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}
4xy^{4}-2y+\dfrac{1}{x^{2}y^1}
\\\\=
4xy^{4}-2y+\dfrac{1}{x^{2}y}
.\end{array}
