1. If $\frac{x}{2 y}=\frac{y}{x-y}$, then $\frac{x^{2}-2 x y-3 y^{2}}{x^{2}+4 x y+3 y^{2}}=(\quad)$.
(A) 0
(B) $-\frac{3}{2}$
(C) $-\frac{1}{5}$
(D) $-\frac{1}{5}$ or $-\frac{3}{2}$
- 1. C.

From $\frac{x}{2 y}=\frac{y}{x-y}$, we get
$$
\begin{array}{l}
x^{2}-x y-2 y^{2}=0(x \neq y \neq 0) \\
\Rightarrow x=-y \text { or } x=2 y . \\
\text { Therefore, } \frac{x^{2}-2 x y-3 y^{2}}{x^{2}+4 x y+3 y^{2}}=\frac{(x-3 y)(x+y)}{(x+3 y)(x+y)} \\
=\frac{x-3 y}{x+3 y}=-\frac{1}{5}(x \neq-y) .
\end{array}
$$