$1 \cdot 5$ If $x, y$ and $2 x+\frac{y}{2}$ are nonzero numbers, then
$$
\left(2 x+\frac{y}{2}\right)^{-1}\left[(2 x)^{-1}+\left(\frac{y}{2}\right)^{-1}\right]
$$

equals
(A) 1 .
(B) $x y^{-1}$.
(C) $x^{-1} y$.
(D) $(x y)^{-1}$.
(E) None of these.
(28th American High School Mathematics Examination, 1977)
[Solution] Notice the following algebraic transformation:
$$
\begin{aligned}
& \left(2 x+\frac{y}{2}\right)^{-1}\left[(2 x)^{-1}+\left(\frac{y}{2}\right)^{-1}\right] \\
= & \frac{1}{2 x+\frac{y}{2}}\left(\frac{1}{2 x}+\frac{2}{y}\right) \\
= & \frac{1}{2 x+\frac{y}{2}} \cdot \frac{\left(\frac{y}{2}+2 x\right)}{x y} \\
= & \frac{1}{x y}=(x y)^{-1} .
\end{aligned}
$$

Therefore, the correct choice is $(D)$.