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