Calculate the term independent of $x$ in the binomial expansion of $\left(x - \frac{1}{2x^5}\right)^{18}$.

1. Write down the general term of the binomial expansion

The general term is given by:
$T_{r+1} = \binom{18}{r} (x)^{18-r} \left(-\frac{1}{2x^5}\right)^r$
$T_{r+1} = \binom{18}{r} x^{18-r} \left(-\frac{1}{2}\right)^r x^{-5r}$
$T_{r+1} = \binom{18}{r} \left(-\frac{1}{2}\right)^r x^{18-6r}$

2. Find the value of $r$ for the term independent of $x$

For the term independent of $x$, the power of $x$ must be $0$:
$18 - 6r = 0$
$6r = 18$
$r = 3$

3. Calculate the term independent of $x$

Substitute $r = 3$ into the general term:
$T_{4} = \binom{18}{3} \left(-\frac{1}{2}\right)^3$
$T_{4} = 816 \times \left(-\frac{1}{8}\right)$
$T_{4} = -102$