在 $\left(x^2 - \frac{m}{2x}\right)^{12}$ 的展开式中，其中 $m$ 是正常数，不含 $x$ 的常数项为 126 720。

1. Write down the general term of the binomial expansion

The general term $T_{r+1}$ in the expansion of $\left(x^2 - \frac{m}{2x}\right)^{12}$ is:
$T_{r+1} = \binom{12}{r} (x^2)^{12-r} \left(-\frac{m}{2x}\right)^r$

2. Simplify the general term to isolate the power of $x$

$T_{r+1} = \binom{12}{r} x^{24-2r} \left(-\frac{m}{2}\right)^r x^{-r}$
$T_{r+1} = \binom{12}{r} \left(-\frac{m}{2}\right)^r x^{24-3r}$

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

For the term independent of $x$, the power of $x$ must be 0:
$24 - 3r = 0$
$3r = 24$
$r = 8$

4. Equate the coefficient to the given value and solve for $m$

Substitute $r = 8$ into the coefficient part:
$\binom{12}{8} \left(-\frac{m}{2}\right)^8 = 126720$
$495 \times \frac{m^8}{256} = 126720$
$\frac{m^8}{256} = \frac{126720}{495}$
$\frac{m^8}{256} = 256$
$m^8 = 256 \times 256 = 65536$
Since $m$ is a positive constant:
$m = \sqrt[8]{65536} = 4$