$6 \cdot 14$ Let $x=1+2^{p}, \quad y=1+2^{-p}$. Then $y$ equals
(A) $\frac{x+1}{x-1}$.
(B) $\frac{x+2}{x-1}$.
(C) $\frac{x}{x-1}$.
(D) $2-x$.
(E) $\frac{x-1}{x}$.
(China Beijing Junior High School Mathematics Competition, 1984)
[Solution] From the given, we have $x-1=2^{p}, y-1=2^{-p}$, so $(x-1)(y-1)=2^{p} \cdot 2^{-p}=1$. Therefore, $y-1=\frac{1}{x-1}$, which means $y=\frac{1}{x-1}+1=\frac{x}{x-1}$. Hence, the answer is $(C)$.