7・105 Let $f(x)=\frac{1+x}{1-3 x}$, and $f_{1}(x)=f[f(x)], \quad f_{2}(x)=$ $f\left[f_{1}(x)\right], \cdots, f_{n}(x)=f\left[f_{n-1}(x)\right], \cdots$, then the value of $f_{1991}(4.7)$ is 
(A) $-\frac{57}{131}$.
(B) $\frac{37}{151}$.
(C) 1991 .
(D) 4.7 .
(China Beijing High School Mathematics Competition, 1991)
[Solution] Given that $f_{1}(x)=\frac{1+\frac{1+x}{1-3 x}}{1-3 \cdot \frac{1+x}{1-3 x}}=\frac{x-1}{1+x}$,
$$
f_{2}(x)=\frac{1+\frac{x-1}{1+x}}{1-3 \cdot \frac{x-1}{1+x}}=x,
$$

thus $f_{3}(x)=f(x), f_{4}(x)=f_{1}(x), f_{5}(x)=f_{2}(x), \cdots$
that is,
$$
f_{3 n+r}(x)=f_{r}(x)(r=0,1,2, \text{ and } n \in N),
$$

where $f_{0}(x)=f(x)$.
Therefore, $f_{1991}(x)=f_{663 \cdot 3+2}(x)=f_{2}(x)=x$,

so
$$
f_{1991}(4.7)=4.7 \text{. }
$$

Hence, the answer is $(D)$.