Find the constant term in the expansion of

$$\\\left(z-\frac{2}{\sqrt{z}}\right)^9\\\\ $The general term is$\\\\ (9Cr)(z)^{(9-r)}\left(\frac{-2}{\sqrt{z}\right)^r}\\\\ =(9Cr)\left(\dfrac{z^9}{z^r}\right)\left(\dfrac{(-2)^r}{(\sqrt{z})^r\right)}\\\\ =(9Cr)\left(\dfrac{z^9*(-2)^r}{z^r*(\sqrt{z})^r\right)}\right)\\\\ =(9Cr)\left(\dfrac{z^{18/2}*(-2)^r}{z^{(2r/2)}*(z)^{(r/2)}\right)}\right)\\\\ =(9Cr)\left(\dfrac{z^{((18-2r-r)/2)}*(-2)^r}{1}\right)\\\\ =(9Cr)(z^{((18-3r)/2)}*(-2)^r)\\\\$$

$$\\$The constant term will be when $\\\\ (18-3r)/2=0\\ 18-3r=0\\ r=6\\ S$so the constant term is $\\ =(9C6)*(-2)^6)\\ =64*9C6\\\\$$

$${\mathtt{64}}{\mathtt{\,\times\,}}{\left({\frac{{\mathtt{9}}{!}}{{\mathtt{6}}{!}{\mathtt{\,\times\,}}({\mathtt{9}}{\mathtt{\,-\,}}{\mathtt{6}}){!}}}\right)} = {\mathtt{5\,376}}$$

You had better check that answer Mellie       

We'll let WolframAlpha do the heavy lifting, here....

Looks like the constant term is 5376....

We could have actually determined this through the binomial expansion

The term would be  C(9,6)z^3 [-2*z^(-1/2)]^6 = C(9,6)[-2]^6 = 5376 .....note that the z's "cancel"

WOW I got the same answer as CPhill