smc

By the Binomial Theorem, each term of the expansion is $\left(\genfrac{}{}{0ex}{}{8}{n}\right){\left(\frac{x^2}{2}\right)}^{8-n}{\left(\frac{-2}{x}\right)}^n$. We want the exponent of $x$ to be $7$, so $$2(8-n)-n=7$$ $$16-3n=7$$ $$n=3$$ If $n=3$, then the corresponding term is $$\left(\genfrac{}{}{0ex}{}{8}{3}\right){\left(\frac{x^2}{2}\right)}^5{\left(\frac{-2}{x}\right)}^3$$ $$56\cdot \frac{x^{10}}{32}\cdot \frac{-8}{x^3}$$ $$-14x^7$$ The answer is $\boxed{-14}$.