(B)
5.【Analysis and Solution】Let $y=\omega x, \omega \neq 1$, substituting into the given condition we get
$$
1+\omega+\omega^{2}=0 \Rightarrow(1-\omega)\left(1+\omega+\omega^{2}\right)=0 \text {, }
$$

thus $\omega^{3}=1$,
$$
\begin{aligned}
\text { hence the original expression } & =\frac{1}{(1+\omega)^{1990}}+\frac{\omega^{1990}}{(1+\omega)^{1990}} \\
& =\frac{1+\omega^{1990}}{\left(1-\omega^{2}\right)^{1990}}=\frac{1+\omega}{\omega^{2}}=-1 .
\end{aligned}
$$
5. Let non-zero complex numbers $x, y$ satisfy $x^{2}+x y+y^{2}=0$, then the value of the algebraic expression $\left(\frac{x}{x+y}\right)^{1990}+\left(\frac{y}{x+y}\right)^{1990}$ is
(A) $2^{-1989}$
(B) -1
(C) 1
(D) None of the above answers is correct