`f(x)=sqrt((x-1)/(x-2{x}))`, where `{*}` denotes the fractional part.

Answer

To find the domain of the function \( f(x) = \sqrt{\frac{x - 1}{x - 2\{x\}}} \), where \(\{x\}\) denotes the fractional part of \(x\), we need to ensure that the expression under the square root is positive and that the denominator is not zero.

### Step 1: Ensure the expression under the square root is positive
The expression under the square root is:
\[ \frac{x - 1}{x - 2\{x\}} > 0 \]
This means both the numerator and denominator must either be both positive or both negative.

#### Sub-step 1.1: Analyze the numerator
The numerator is \(x - 1\):
- \(x - 1 > 0 \implies x > 1\)

#### Sub-step 1.2: Analyze the denominator
The denominator is \(x - 2\{x\}\). We know that the fractional part \(\{x\}\) is defined as:
\[ \{x\} = x - \lfloor x \rfloor \]
This means \(0 \leq \{x\} < 1\). Therefore, \(2\{x\}\) will range from \(0\) to \(2\).

The denominator can be rewritten as:
\[ x - 2\{x\} = x - 2(x - \lfloor x \rfloor) = \lfloor x \rfloor - x + 2\lfloor x \rfloor = 2\lfloor x \rfloor - x \]
For the denominator to be positive:
\[ x - 2\{x\} > 0 \implies x > 2\{x\} \]
Since \(2\{x\} < 2\), this condition will hold for \(x > 2\).

### Step 2: Ensure the denominator is not zero
We also need to ensure that the denominator does not equal zero:
\[ x - 2\{x\} \neq 0 \]
This occurs when:
\[ x \neq 2\{x\} \]
Since \(\{x\} < 1\), \(2\{x\}\) will always be less than \(2\). Therefore, \(x\) cannot equal \(2\{x\}\) for any \(x\) in the domain.

### Step 3: Combine conditions
From the analysis:
1. \(x > 1\) from the numerator.
2. \(x > 2\) from the denominator.
3. \(x \neq 0\) since the denominator must not be zero.

### Conclusion
The domain of \(f(x)\) is:
\[ \text{Domain} = (2, \infty) \]