In the expansion of $\left(a-\dfrac{1}{\sqrt{a}}\right)^7$ the coefficient of $a^{-\dfrac{1}{2}}$ is:
$\textbf{(A)}\ -7 \qquad \textbf{(B)}\ 7 \qquad \textbf{(C)}\ -21 \qquad \textbf{(D)}\ 21 \qquad \textbf{(E)}\ 35$
By the [Binomial Theorem](https://artofproblemsolving.com/wiki/index.php/Binomial_Theorem), each term of the expansion is $\binom{7}{n}(a)^{7-n}(\frac{-1}{\sqrt{a}})^n$.
We want the exponent of $a$ to be $-\frac{1}{2}$, so 
\[(7-n)-\frac{1}{2}n=-\frac{1}{2}\]
\[-\frac{3}{2}n = -\frac{15}{2}\]
\[n = 5\]
If $n=5$, then the corresponding term is
\[\binom{7}{5}(a)^{2}(\frac{-1}{\sqrt{a}})^5\]
\[-21a^{-\frac{1}{2}}\]
The answer is $\boxed{\textbf{(C)}}$.