Given the linear fractional transformation of $x$ into $f_1(x)=\dfrac{2x-1}{x+1}$. Define $f_{n+1}(x)=f_1(f_n(x))$ for $n=1,2,3,\cdots$. Assuming that $f_{35}(x)=f_5(x)$, it follows that $f_{28}(x)$ is equal to

$\textbf{(A) }x\qquad\textbf{(B) }\frac{1}{x}\qquad\textbf{(C) }\frac{x-1}{x}\qquad\textbf{(D) }\frac{1}{1-x}\qquad \textbf{(E) }\text{None of these}$
1. **Identify the periodicity of the function:**
   We start by computing the first few iterations of the function \( f_1(x) = \frac{2x-1}{x+1} \) to identify any periodicity.

   \[
   f_2(x) = f_1(f_1(x)) = f_1\left(\frac{2x-1}{x+1}\right)
   \]

   To compute \( f_2(x) \):

   \[
   f_2(x) = \frac{2\left(\frac{2x-1}{x+1}\right) - 1}{\left(\frac{2x-1}{x+1}\right) + 1} = \frac{\frac{4x-2 - (x+1)}{x+1}}{\frac{2x-1 + (x+1)}{x+1}} = \frac{\frac{4x-2-x-1}{x+1}}{\frac{2x-1+x+1}{x+1}} = \frac{\frac{3x-3}{x+1}}{\frac{3x}{x+1}} = \frac{3(x-1)}{3x} = \frac{x-1}{x}
   \]

   \[
   f_2(x) = \frac{x-1}{x}
   \]

2. **Compute further iterations to find the periodicity:**

   \[
   f_3(x) = f_1(f_2(x)) = f_1\left(\frac{x-1}{x}\right)
   \]

   To compute \( f_3(x) \):

   \[
   f_3(x) = \frac{2\left(\frac{x-1}{x}\right) - 1}{\left(\frac{x-1}{x}\right) + 1} = \frac{\frac{2(x-1) - x}{x}}{\frac{x-1 + x}{x}} = \frac{\frac{2x-2-x}{x}}{\frac{2x-1}{x}} = \frac{\frac{x-2}{x}}{\frac{2x-1}{x}} = \frac{x-2}{2x-1}
   \]

3. **Continue to find \( f_4(x) \):**

   \[
   f_4(x) = f_1(f_3(x)) = f_1\left(\frac{x-2}{2x-1}\right)
   \]

   To compute \( f_4(x) \):

   \[
   f_4(x) = \frac{2\left(\frac{x-2}{2x-1}\right) - 1}{\left(\frac{x-2}{2x-1}\right) + 1} = \frac{\frac{2(x-2) - (2x-1)}{2x-1}}{\frac{x-2 + (2x-1)}{2x-1}} = \frac{\frac{2x-4 - 2x + 1}{2x-1}}{\frac{3x-3}{2x-1}} = \frac{-3}{3x-3} = -\frac{1}{x-1}
   \]

4. **Compute \( f_5(x) \):**

   \[
   f_5(x) = f_1(f_4(x)) = f_1\left(-\frac{1}{x-1}\right)
   \]

   To compute \( f_5(x) \):

   \[
   f_5(x) = \frac{2\left(-\frac{1}{x-1}\right) - 1}{\left(-\frac{1}{x-1}\right) + 1} = \frac{\frac{-2 - (x-1)}{x-1}}{\frac{-1 + (x-1)}{x-1}} = \frac{\frac{-2 - x + 1}{x-1}}{\frac{x-2}{x-1}} = \frac{-x-1}{x-2} = -\frac{x+1}{x-2}
   \]

5. **Compute \( f_6(x) \):**

   \[
   f_6(x) = f_1(f_5(x)) = f_1\left(-\frac{x+1}{x-2}\right)
   \]

   To compute \( f_6(x) \):

   \[
   f_6(x) = \frac{2\left(-\frac{x+1}{x-2}\right) - 1}{\left(-\frac{x+1}{x-2}\right) + 1} = \frac{\frac{-2(x+1) - (x-2)}{x-2}}{\frac{-(x+1) + (x-2)}{x-2}} = \frac{\frac{-2x-2 - x + 2}{x-2}}{\frac{-x-1 + x - 2}{x-2}} = \frac{\frac{-3x}{x-2}}{\frac{-3}{x-2}} = x
   \]

   We observe that \( f_6(x) = x \), indicating that the sequence is periodic with a period of 6.

6. **Use the periodicity to find \( f_{28}(x) \):**

   Since the sequence is periodic with a period of 6, we have:

   \[
   f_{28}(x) = f_{28 \mod 6}(x) = f_4(x)
   \]

   From our previous computation, we know:

   \[
   f_4(x) = -\frac{1}{x-1}
   \]

   Therefore, \( f_{28}(x) = -\frac{1}{x-1} \).

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