The solution set of the inequation `|(2x-1)/(x-1)| gt 2`, is

Options

A. `(34, 1) cup (1, oo)`
B. `(3//4, oo)`
C. `(-oo, 3//4)`
D. none of these

Answer

To solve the inequality \( \left| \frac{2x-1}{x-1} \right| > 2 \), we will break it down into two separate cases based on the properties of absolute values.

### Step 1: Set up the inequality
The inequality can be expressed as two separate inequalities:
1. \( \frac{2x-1}{x-1} > 2 \)
2. \( \frac{2x-1}{x-1} < -2 \)

### Step 2: Solve the first inequality
**Inequality 1:**
\[ \frac{2x-1}{x-1} > 2 \]

Multiply both sides by \( x - 1 \) (noting that we need to consider the sign of \( x - 1 \)):
- If \( x - 1 > 0 \) (i.e., \( x > 1 \)):
\[ 2x - 1 > 2(x - 1) \implies 2x - 1 > 2x - 2 \implies -1 > -2 \quad \text{(Always true)} \]
Thus, for \( x > 1 \), the inequality holds.

- If \( x - 1 < 0 \) (i.e., \( x < 1 \)):
\[ 2x - 1 < 2(x - 1) \implies 2x - 1 < 2x - 2 \implies -1 < -2 \quad \text{(Never true)} \]
Thus, there are no solutions for \( x < 1 \).

### Step 3: Solve the second inequality
**Inequality 2:**
\[ \frac{2x-1}{x-1} < -2 \]

Again, multiply both sides by \( x - 1 \):
- If \( x - 1 > 0 \) (i.e., \( x > 1 \)):
\[ 2x - 1 < -2(x - 1) \implies 2x - 1 < -2x + 2 \implies 4x < 3 \implies x < \frac{3}{4} \]
Since \( x > 1 \) and \( x < \frac{3}{4} \) cannot both be true, there are no solutions in this case.

- If \( x - 1 < 0 \) (i.e., \( x < 1 \)):
\[ 2x - 1 > -2(x - 1) \implies 2x - 1 > -2x + 2 \implies 4x > 3 \implies x > \frac{3}{4} \]
Thus, for \( x < 1 \), we have the solution \( \frac{3}{4} < x < 1 \).

### Step 4: Combine the solutions
From the two inequalities, we have:
1. From the first inequality: \( x > 1 \)
2. From the second inequality: \( \frac{3}{4} < x < 1 \)

The overall solution set is the union of these intervals:
\[ x \in \left( \frac{3}{4}, 1 \right) \cup (1, \infty) \]

### Final Answer
The solution set of the inequation \( \left| \frac{2x-1}{x-1} \right| > 2 \) is:
\[ \left( \frac{3}{4}, 1 \right) \cup (1, \infty) \]