To solve the equation \(\frac{x-5}{x-1} + \frac{2}{x-3} = 1\), we need to find a common denominator and then solve for \(x\). Here are the steps:

1. **Find a common denominator:**
The common denominator for the fractions \(\frac{x-5}{x-1}\) and \(\frac{2}{x-3}\) is \((x-1)(x-3)\).

2. **Rewrite each fraction with the common denominator:**
\[
\frac{x-5}{x-1} = \frac{(x-5)(x-3)}{(x-1)(x-3)}
\]
\[
\frac{2}{x-3} = \frac{2(x-1)}{(x-1)(x-3)}
\]

3. **Combine the fractions:**
\[
\frac{(x-5)(x-3) + 2(x-1)}{(x-1)(x-3)} = 1
\]

4. **Simplify the numerator:**
\[
(x-5)(x-3) + 2(x-1) = (x^2 - 3x - 5x + 15) + (2x - 2) = x^2 - 8x + 15 + 2x - 2 = x^2 - 6x + 13
\]

5. **Set the equation equal to 1:**
\[
\frac{x^2 - 6x + 13}{(x-1)(x-3)} = 1
\]

6. **Clear the fraction by multiplying both sides by \((x-1)(x-3)\):**
\[
x^2 - 6x + 13 = (x-1)(x-3)
\]

7. **Expand the right side:**
\[
x^2 - 6x + 13 = x^2 - 4x + 3
\]

8. **Subtract \(x^2 - 4x + 3\) from both sides:**
\[
x^2 - 6x + 13 - (x^2 - 4x + 3) = 0
\]
\[
x^2 - 6x + 13 - x^2 + 4x - 3 = 0
\]
\[
-2x + 10 = 0
\]

9. **Solve for \(x\):**
\[
-2x + 10 = 0
\]
\[
-2x = -10
\]
\[
x = 5
\]

10. **Check for extraneous solutions:**
Substitute \(x = 5\) back into the original equation to ensure it does not make any denominator zero:
\[
\frac{5-5}{5-1} + \frac{2}{5-3} = \frac{0}{4} + \frac{2}{2} = 0 + 1 = 1
\]

Since \(x = 5\) does not make any denominator zero and satisfies the original equation, the solution is:
\[
\boxed{5}
\]