7. If a non-zero complex number $x$ satisfies $x+\frac{1}{x}=1$, then $x^{2014}+\frac{1}{x^{2014}}=$ $\qquad$ .
II.7. -1.
From $x+\frac{1}{x}=1 \Rightarrow x=\mathrm{e}^{ \pm \frac{x}{3} \mathrm{i}}$.
Therefore, $x^{2014}=\mathrm{e}^{\mathrm{E}^{\frac{2014 \pi}{3}} \mathrm{i}}=\mathrm{e}^{ \pm \frac{4 \pi}{3} \mathrm{i}}$.
Thus, $x^{2014}+\frac{1}{x^{2014}}=2 \cos \frac{4 \pi}{3}=-1$.