Solution:

The answer is (C). It is possible to explicitly determine the values of the solutions of the equation. This is in fact a reciprocal equation of the first kind (equations in which the coefficient of $x^{k}$ is equal to that of $x^{n-k}$, where $n$ is the degree of the equation and $k=0,1, \ldots, n$) and they have the characteristic that if they are satisfied by a certain value $x_{0}$ they are also satisfied by $1 / x_{0}$, hence the name "reciprocal". The solution of equations of this type can be reduced to that of three second-degree equations: the equation can be written in the form $x^{2}\left[\left(x^{2}+\frac{1}{x^{2}}\right)-2\left(x+\frac{1}{x}\right)-7\right]=0$. Setting $y=x+\frac{1}{x}$ we arrive at $\left(y^{2}-2\right)-2 y-7=0$ from which we determine two values of $y$ and four of $x$. This way of proceeding, however, leads to rather laborious calculations: the values of the solutions $x_{1}, x_{2}, x_{3}$ and $x_{4}$ (and hence also of their reciprocals) are in fact
$$x_{1,2}=\frac{1+\sqrt{10}}{2} \pm \frac{\sqrt{5}+\sqrt{2}}{2} \quad \text{e} \quad x_{3,4}=\frac{1-\sqrt{10}}{2} \pm \frac{\sqrt{5}-\sqrt{2}}{2}.$$
pairwise reciprocal to each other, so that
$$\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}=x_{2}+x_{1}+x_{4}+x_{3}=1+\sqrt{10}+1-\sqrt{10}=2.$$

One arrives much more quickly at the result by recalling that if $x_{1}, x_{2}, x_{3}$ and $x_{4}$ are solutions of the equation, then $x^{4}-2 x^{3}-7 x^{2}-2 x+1=\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)$. Carrying out the calculations on the right-hand side we note that the coefficient of the third degree of the product is $-\left(x_{1}+x_{2}+x_{3}+x_{4}\right)$ which, by comparison with the left-hand side, is equal to $-2$. Since the equation is reciprocal we can immediately conclude that $\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}=2$.

Finally, note that it is possible to solve the exercise even without taking into account that the equation is reciprocal: again computing the product $\left(x-x_{1}\right)\left(x-x_{2}\right)\left(x-x_{3}\right)\left(x-x_{4}\right)$ we observe that
$$\frac{1}{x_{1}}+\frac{1}{x_{2}}+\frac{1}{x_{3}}+\frac{1}{x_{4}}=\frac{x_{1} x_{2} x_{3}+x_{1} x_{2} x_{4}+x_{1} x_{3} x_{4}+x_{2} x_{3} x_{4}}{x_{1} x_{2} x_{3} x_{4}}=\frac{- \text{coeff of first degree}}{\text{constant term}}=\frac{2}{1}.$$