How do I solve this problem using binomial theorem with square root in it?

Question

The question is: Find the coefficient of $\frac{1}{x\sqrt x}$ in the expansion of $(x^2 - \frac{1}{2\sqrt x})^{18}$.

I have included a photo to make it easier to read because I do not know how to format the question.

Answer 1

Hint:

Any term in the expansion would be of the form $${18\choose r} \cdot (x^2)^r \cdot\left(\frac 12 x^{-1/2}\right)^{18-r} $$ You need $x^{2r -\frac{18-r}{2}}=\frac{1}{x\sqrt x} =x^{-3/2}$ or $$2r-\frac{18-r}{2} =-\frac 32$$ Solve for $r$ from this equation and just plug its value in the first expression.

Answer 2

Hint:

Use the binomial formula: $$\biggl(x^2-\frac 1{2\sqrt x}\biggr)^{18}=\sum_{k=0}^{18}\binom{18}k x^{2k}\frac1{2^{18-k}x^{\tfrac{18-k}2}}$$ and determine $k$ such that $\;2k-\frac{18-k}2=-\frac 32$