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.
5. (B)

Let $y=\omega x, \omega \neq 1$, substituting into the given condition we get
$$
1+\omega+\omega^{2}=0 \Rightarrow(1-\omega)\left(1+\omega+\omega^{2}\right)=0
$$

Thus,
$$
\omega^{3}=1,
$$

Therefore,
$$
\begin{array}{l}
\text { Hence } \quad \text { the original expression }=\frac{1}{(1+\omega)^{1990}}+\frac{\omega^{1990}}{(1+\omega)^{1990}} \\
=\frac{1+\omega^{1990}}{\left(1-\omega^{2}\right)^{1990}}=\frac{1+\omega}{\omega^{2}}=-1 .
\end{array}
$$