If $x, y$ and $2x + \frac{y}{2}$ are not zero, then
$\left( 2x + \frac{y}{2} \right)^{-1} \left[(2x)^{-1} + \left( \frac{y}{2} \right)^{-1} \right]$
equals
$\textbf{(A) }1\qquad \textbf{(B) }xy^{-1}\qquad \textbf{(C) }x^{-1}y\qquad \textbf{(D) }(xy)^{-1}\qquad  \textbf{(E) }\text{none of these}$
We can write $\left( 2x+ \frac{y}{2} \right)^{-1} = \left( \frac{4x+y}{2} \right)^{-1} = \frac{2}{4x+y}$.
Then, the expression simplifies to \[\left( 2x + \frac{y}{2} \right)^{-1} \left[(2x)^{-1} + \left( \frac{y}{2} \right)^{-1} \right] \Rightarrow \frac{2}{4x+y} \left( \frac{1}{2x} + \frac{2}{y} \right) \Rightarrow \frac{2}{4x+y} \left( \frac{y+4x}{2xy} \right) \Rightarrow \frac{1}{xy}.\]

Thus, our answer is $\textbf{(D) }(xy)^{-1}$. ~[jiang147369](https://artofproblemsolving.com/wiki/index.php/User:Jiang147369)