Answer
$\color{blue}{\dfrac{1}{2k^2(k-1)}}$
Work Step by Step
Factor each polynomial to obtain:
$=\dfrac{k(k+1)}{4k(2k^2)} \cdot \dfrac{4}{(k+1)(k-1)}$
Cancel the common factors then multiply:
$\require{cancel}
\\=\dfrac{\cancel{k}\cancel{(k+1)}}{\cancel{4}\cancel{k}(2k^2)} \cdot \dfrac{\cancel{4}}{\cancel{(k+1)}(k-1)}
\\=\dfrac{1}{2k^2}\cdot \dfrac{1}{k-1}
\\=\color{blue}{\dfrac{1}{2k^2(k-1)}}$
