Expand using Binomial Theorem $\left(1+\frac{x}{2}-\frac{2}{x}\right)^{4}, x \neq 0$.

Let $a = 1 + \frac{x}{2}$ and $b = \frac{2}{x}$. Then the expression is $(a - b)^4$.
Using the Binomial Theorem,$(a - b)^4 = a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4$.
Substituting $a$ and $b$:
$= (1 + \frac{x}{2})^4 - 4(1 + \frac{x}{2})^3(\frac{2}{x}) + 6(1 + \frac{x}{2})^2(\frac{2}{x})^2 - 4(1 + \frac{x}{2})(\frac{2}{x})^3 + (\frac{2}{x})^4$
$= (1 + \frac{x}{2})^4 - \frac{8}{x}(1 + \frac{x}{2})^3 + \frac{24}{x^2}(1 + x + \frac{x^2}{4}) - \frac{32}{x^3}(1 + \frac{x}{2}) + \frac{16}{x^4}$
$= (1 + \frac{x}{2})^4 - \frac{8}{x}(1 + \frac{x}{2})^3 + \frac{24}{x^2} + \frac{24}{x} + 6 - \frac{32}{x^3} - \frac{16}{x^2} + \frac{16}{x^4}$
$= (1 + \frac{x}{2})^4 - \frac{8}{x}(1 + \frac{x}{2})^3 + \frac{8}{x^2} + \frac{24}{x} + 6 - \frac{32}{x^3} + \frac{16}{x^4} \quad \dots(1)$
Expanding $(1 + \frac{x}{2})^4 = 1 + 4(\frac{x}{2}) + 6(\frac{x^2}{4}) + 4(\frac{x^3}{8}) + \frac{x^4}{16} = 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} \quad \dots(2)$
Expanding $(1 + \frac{x}{2})^3 = 1 + 3(\frac{x}{2}) + 3(\frac{x^2}{4}) + \frac{x^3}{8} = 1 + \frac{3x}{2} + \frac{3x^2}{4} + \frac{x^3}{8} \quad \dots(3)$
Substituting $(2)$ and $(3)$ into $(1)$:
$= (1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16}) - \frac{8}{x}(1 + \frac{3x}{2} + \frac{3x^2}{4} + \frac{x^3}{8}) + \frac{8}{x^2} + \frac{24}{x} + 6 - \frac{32}{x^3} + \frac{16}{x^4}$
$= 1 + 2x + \frac{3x^2}{2} + \frac{x^3}{2} + \frac{x^4}{16} - \frac{8}{x} - 12 - 6x - x^2 + \frac{8}{x^2} + \frac{24}{x} + 6 - \frac{32}{x^3} + \frac{16}{x^4}$
$= \frac{16}{x^4} - \frac{32}{x^3} + \frac{8}{x^2} + \frac{16}{x} + \frac{x^4}{16} + \frac{x^3}{2} + \frac{x^2}{2} - 4x - 5$.