$8 \cdot 31$ If $x>0, \quad y>0$, and $2 \lg (x-2 y)=\lg x+\lg y$, then the value of $x: y$ is
(A) 4 .
(B) 1 .
(C) 1 or 4 .
(D) $\frac{1}{4}$.
(Tianjin, China Junior High School Mathematics Competition, 1983)
[Solution] Given $\lg (x-2 y)^{2}=\lg x y$, then $(x-2 y)^{2}=x y$, which means $x^{2}-5 x y+4 y^{2}=0$. By $y>0$, the equation can be transformed into $\left(\frac{x}{y}\right)^{2}-5\left(\frac{x}{y}\right)+4=0$, solving this yields $\frac{x}{y}=4$ or 1. However, $\frac{x}{y}=1$, i.e., $x=y$ leads to $x-2 y<0$, which is not valid, so it is discarded. In summary, $x: y=4$, hence the answer is $(A)$.