Expand the expression

By using Binomial Theorem, the expression $\left(\frac{2}{x}-\frac{x}{2}\right)^{5}$ can be expanded as

$\left(\frac{2}{x}-\frac{x}{2}\right)^{5}={ }^{5} C_{0}\left(\frac{2}{x}\right)^{5}-{ }^{5} C_{1}\left(\frac{2}{x}\right)^{4}\left(\frac{x}{2}\right)+{ }^{5} C_{2}\left(\frac{2}{x}\right)^{3}\left(\frac{x}{2}\right)^{2}$

$-{ }^{5} C_{3}\left(\frac{2}{x}\right)^{2}\left(\frac{x}{2}\right)^{3}+{ }^{5} C_{4}\left(\frac{2}{x}\right)\left(\frac{x}{2}\right)^{4}-{ }^{5} C_{5}\left(\frac{x}{2}\right)^{5}$

$=\frac{32}{x^{5}}-5\left(\frac{16}{x^{4}}\right)\left(\frac{x}{2}\right)+10\left(\frac{8}{x^{3}}\right)\left(\frac{x^{2}}{4}\right)-10\left(\frac{4}{x^{2}}\right)\left(\frac{x^{3}}{8}\right)+5\left(\frac{2}{x}\right)\left(\frac{x^{4}}{16}\right)-\frac{x^{5}}{32}$

$=\frac{32}{x^{5}}-\frac{40}{x^{3}}+\frac{20}{x}-5 x+\frac{5}{8} x^{3}-\frac{x^{3}}{32}$