To solve the inequality \(\frac{|x-1|}{x-2} > 0\), we will analyze the expression step by step.
### Step 1: Understand the Absolute Value
The expression involves an absolute value, which we need to break down into cases based on the value of \(x\).
- **Case 1:** When \(x - 1 \geq 0\) (i.e., \(x \geq 1\)), then \(|x - 1| = x - 1\).
- **Case 2:** When \(x - 1 < 0\) (i.e., \(x < 1\)), then \(|x - 1| = -(x - 1) = 1 - x\).
### Step 2: Analyze Case 1 (\(x \geq 1\))
For \(x \geq 1\), the inequality becomes:
\[
\frac{x - 1}{x - 2} > 0
\]
This fraction is positive when both the numerator and denominator are either both positive or both negative.
1. **Numerator:** \(x - 1 > 0 \Rightarrow x > 1\)
2. **Denominator:** \(x - 2 > 0 \Rightarrow x > 2\)
Now, we find the critical points:
- \(x = 1\) (numerator is zero)
- \(x = 2\) (denominator is zero)
### Step 3: Test Intervals for Case 1
We will test the intervals based on the critical points \(1\) and \(2\):
- **Interval 1:** \( (-\infty, 1) \)
- **Interval 2:** \( (1, 2) \)
- **Interval 3:** \( (2, \infty) \)
Testing these intervals:
- For \(x = 0\) (in \((- \infty, 1)\)):
\[
\frac{0 - 1}{0 - 2} = \frac{-1}{-2} = \frac{1}{2} > 0 \quad \text{(positive)}
\]
- For \(x = 1.5\) (in \((1, 2)\)):
\[
\frac{1.5 - 1}{1.5 - 2} = \frac{0.5}{-0.5} = -1 < 0 \quad \text{(negative)}
\]
- For \(x = 3\) (in \((2, \infty)\)):
\[
\frac{3 - 1}{3 - 2} = \frac{2}{1} = 2 > 0 \quad \text{(positive)}
\]
### Step 4: Conclusion for Case 1
From the testing:
- The inequality is satisfied in the intervals \((- \infty, 1)\) and \((2, \infty)\).
### Step 5: Analyze Case 2 (\(x < 1\))
For \(x < 1\), the inequality becomes:
\[
\frac{1 - x}{x - 2} > 0
\]
Again, we analyze when both the numerator and denominator are positive or both are negative.
1. **Numerator:** \(1 - x > 0 \Rightarrow x < 1\) (always true in this case)
2. **Denominator:** \(x - 2 < 0 \Rightarrow x < 2\) (always true since \(x < 1\))
### Step 6: Test Intervals for Case 2
We only need to check if this inequality holds for \(x < 1\):
- For \(x = 0\):
\[
\frac{1 - 0}{0 - 2} = \frac{1}{-2} < 0 \quad \text{(negative)}
\]
### Step 7: Conclusion for Case 2
The inequality is not satisfied for any \(x < 1\).
### Final Solution
Combining both cases:
- From Case 1, the solution is \( (-\infty, 1) \cup (2, \infty) \).
- From Case 2, there are no valid solutions.
Thus, the solution set is:
\[
\boxed{(2, \infty)}
\]
