Answer
$\dfrac{1}{2x-1}$
Work Step by Step
The given expression, $
\dfrac{2+\dfrac{1}{x}}{4x-\dfrac{1}{x}}
,$ simplifies to
\begin{array}{l}\require{cancel}
\dfrac{\dfrac{2x+1}{x}}{\dfrac{4x^2-1}{x}}
\\\\=
\dfrac{\dfrac{2x+1}{\cancel{x}}}{\dfrac{4x^2-1}{\cancel{x}}}
\\\\=
\dfrac{2x+1}{4x^2-1}
\\\\=
\dfrac{2x+1}{(2x+1)(2x-1)}
\\\\=
\dfrac{\cancel{2x+1}}{(\cancel{2x+1})(2x-1)}
\\\\=
\dfrac{1}{2x-1}
.\end{array}
