Expand the expression $\left(\frac{2}{x}-\frac{x}{2}\right)^{5}$

Using the Binomial Theorem,the expansion of $\left(\frac{2}{x}-\frac{x}{2}\right)^{5}$ is given by:
$\left(\frac{2}{x}-\frac{x}{2}\right)^{5} = \sum_{k=0}^{5} \binom{5}{k} \left(\frac{2}{x}\right)^{5-k} \left(-\frac{x}{2}\right)^{k}$
$= \binom{5}{0}\left(\frac{2}{x}\right)^{5} - \binom{5}{1}\left(\frac{2}{x}\right)^{4}\left(\frac{x}{2}\right) + \binom{5}{2}\left(\frac{2}{x}\right)^{3}\left(\frac{x}{2}\right)^{2} - \binom{5}{3}\left(\frac{2}{x}\right)^{2}\left(\frac{x}{2}\right)^{3} + \binom{5}{4}\left(\frac{2}{x}\right)\left(\frac{x}{2}\right)^{4} - \binom{5}{5}\left(\frac{x}{2}\right)^{5}$
$= 1 \cdot \frac{32}{x^{5}} - 5 \cdot \frac{16}{x^{4}} \cdot \frac{x}{2} + 10 \cdot \frac{8}{x^{3}} \cdot \frac{x^{2}}{4} - 10 \cdot \frac{4}{x^{2}} \cdot \frac{x^{3}}{8} + 5 \cdot \frac{2}{x} \cdot \frac{x^{4}}{16} - 1 \cdot \frac{x^{5}}{32}$
$= \frac{32}{x^{5}} - \frac{40}{x^{3}} + \frac{20}{x} - 5x + \frac{5x^{3}}{8} - \frac{x^{5}}{32}$