求 $\left(2 - \dfrac{3}{x^{2}}\right)^{7}$ 按 $x$ 的降幂展开式的前 4 项，并化简每一项。 由此，求 $(5x - 4x^{3})\left(2 - \dfrac{3}{x^{2}}\right)^{7}$ 展开式中 $x$ 的系数。 已知 $\left(x^{2} - \frac{k}{x}\right)^{9}$ 的二项式展开式中的常数项为 61236，求正常数 $k$ 的值。

1. Apply the binomial theorem to write down the first 4 terms of the expansion.

$\left(2 - \frac{3}{x^2}\right)^7 = \binom{7}{0}(2)^7 + \binom{7}{1}(2)^6\left(-\frac{3}{x^2}\right) + \binom{7}{2}(2)^5\left(-\frac{3}{x^2}\right)^2 + \binom{7}{3}(2)^4\left(-\frac{3}{x^2}\right)^3 + \dots$

2. Evaluate the binomial coefficients and powers.

$= 1(128) + 7(64)\left(-\frac{3}{x^2}\right) + 21(32)\left(\frac{9}{x^4}\right) + 35(16)\left(-\frac{27}{x^6}\right) + \dots$

3. Simplify each term to obtain the final expansion.

$= 128 - \frac{1344}{x^2} + \frac{6048}{x^4} - \frac{15120}{x^6}$

1. Write down the general term of the expansion

The general term $T_{r+1}$ is given by:
$T_{r+1} = \binom{9}{r} (x^2)^{9-r} \left(-\frac{k}{x}\right)^r$

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

$T_{r+1} = \binom{9}{r} x^{18-2r} (-k)^r x^{-r}$
$T_{r+1} = \binom{9}{r} (-k)^r x^{18-3r}$

3. Find the value of $r$ for the constant term

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

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

Substitute $r = 6$ into the coefficient part:
$\binom{9}{6} (-k)^6 = 61236$
$84 k^6 = 61236$
$k^6 = 729$
Since $k$ is a positive constant:
$k = \sqrt[6]{729} = 3$