The coefficient of $x^7$ in the expansion of $\left(\frac{x^2}{2}-\frac{2}{x}\right)^8$ is:
$\textbf{(A)}\ 56\qquad \textbf{(B)}\ -56\qquad \textbf{(C)}\ 14\qquad \textbf{(D)}\ -14\qquad \textbf{(E)}\ 0$
By the [Binomial Theorem](https://artofproblemsolving.com/wiki/index.php/Binomial_Theorem), each term of the expansion is $\binom{8}{n}\left(\frac{x^2}{2}\right)^{8-n}\left(\frac{-2}{x}\right)^n$.
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
\[\binom{8}{3}\left(\frac{x^2}{2}\right)^{5}\left(\frac{-2}{x}\right)^3\]
\[56 \cdot \frac{x^{10}}{32} \cdot \frac{-8}{x^3}\]
\[-14x^7\]
The answer is $\boxed{\textbf{(D)}}$.