Given the linear fractional transformation of $x$ into $f_1(x) = \tfrac{2x-1}{x+1},$ define $f_{n+1}(x) = f_1(f_n(x))$ for $n=1,2,3,\ldots.$ It can be shown that $f_{35} = f_5.$ Determine $A,B,C,D$ so that $f_{28}(x) = \tfrac{Ax+B}{Cx+D}.$
1. **Understanding the Problem:**
   We are given a linear fractional transformation \( f_1(x) = \frac{2x-1}{x+1} \) and a sequence of functions defined by \( f_{n+1}(x) = f_1(f_n(x)) \). We need to determine the coefficients \( A, B, C, D \) such that \( f_{28}(x) = \frac{Ax + B}{Cx + D} \).

2. **Identifying the Periodicity:**
   It is given that \( f_{35} = f_5 \). This implies that the function has a periodicity of 30, i.e., \( f_{35} = f_5 \) implies \( f^{35} = f^5 \) and hence \( f^{30} = \text{id} \), where \(\text{id}\) is the identity function.

3. **Finding \( f^{-1}(x) \):**
   To find \( f^{-2}(x) \), we first need to determine \( f^{-1}(x) \). The inverse of \( f_1(x) = \frac{2x-1}{x+1} \) can be found by solving:
   \[
   y = \frac{2x-1}{x+1}
   \]
   for \( x \) in terms of \( y \):
   \[
   y(x+1) = 2x - 1 \implies yx + y = 2x - 1 \implies yx - 2x = -1 - y \implies x(y-2) = -1 - y \implies x = \frac{-1 - y}{y - 2}
   \]
   Therefore, 
   \[
   f^{-1}(x) = \frac{-x - 1}{x - 2}
   \]

4. **Finding \( f^{-2}(x) \):**
   Next, we need to find \( f^{-2}(x) \), which is \( f^{-1}(f^{-1}(x)) \):
   \[
   f^{-2}(x) = f^{-1}\left( \frac{-x - 1}{x - 2} \right)
   \]
   Substituting \( \frac{-x - 1}{x - 2} \) into \( f^{-1}(x) \):
   \[
   f^{-1}\left( \frac{-x - 1}{x - 2} \right) = \frac{-\left( \frac{-x - 1}{x - 2} \right) - 1}{\left( \frac{-x - 1}{x - 2} \right) - 2}
   \]
   Simplifying the numerator:
   \[
   -\left( \frac{-x - 1}{x - 2} \right) - 1 = \frac{x + 1}{x - 2} - 1 = \frac{x + 1 - (x - 2)}{x - 2} = \frac{x + 1 - x + 2}{x - 2} = \frac{3}{x - 2}
   \]
   Simplifying the denominator:
   \[
   \left( \frac{-x - 1}{x - 2} \right) - 2 = \frac{-x - 1 - 2(x - 2)}{x - 2} = \frac{-x - 1 - 2x + 4}{x - 2} = \frac{-3x + 3}{x - 2} = \frac{3(1 - x)}{x - 2}
   \]
   Therefore,
   \[
   f^{-2}(x) = \frac{\frac{3}{x - 2}}{\frac{3(1 - x)}{x - 2}} = \frac{3}{3(1 - x)} = \frac{1}{1 - x}
   \]

5. **Conclusion:**
   We have found that \( f_{28}(x) = f^{-2}(x) = \frac{1}{1 - x} \). Therefore, the coefficients \( A, B, C, D \) are:
   \[
   A = 1, \quad B = 0, \quad C = -1, \quad D = 1
   \]

The final answer is \( \boxed{ \frac{1}{1 - x} } \)