3. Let $f_{1}(x)=\frac{2 x-1}{x+1}, f_{n+1}(x)=f_{1}\left[f_{n}(x)\right]$. Then $f_{28}(x)$ equals
A. $x$
B. $\frac{1}{x}$
C. $\frac{x-1}{x}$
D. $\frac{1}{1-x}$
3. $\mathrm{D}$ Hint: $f_{2}(x)=\frac{x-1}{x}, f_{3}(x)=\frac{x-2}{2 x-1}, f_{4}(x)=\frac{1}{1-x}, f_{5}(x)=\frac{x+1}{2-x}, f_{6}(x)=x$.
$$
f_{7}(x)=f_{1}(x), \cdots, f_{28}(x)=f_{4}(x)=\frac{1}{1-x}
$$