7. Given that $a$ is a root of the equation $x^{2}-3 x+1=0$. Then the value of the fraction $\frac{2 a^{6}-6 a^{4}+2 a^{5}-a^{2}-1}{3 a}$ is $\qquad$ -
II. 7. -1.
According to the problem, we have $a^{2}-3 a+1=0$.
$$
\begin{array}{l}
\text { Original expression }=\frac{2 a^{3}\left(a^{2}-3 a+1\right)-\left(a^{2}+1\right)}{3 a} \\
=-\frac{a^{2}+1}{3 a}=-1 .
\end{array}
$$