Find the values of x which the function `f(x)=sqrt(log_(1//2)((x-1)/(x+5))` is defined.

Explanation

To find the values of \( x \) for which the function \( f(x) = \sqrt{\log_{1/2} \left( \frac{x-1}{x+5} \right)} \) is defined, we need to ensure two conditions are satisfied:

1. The argument of the logarithm must be positive.
2. The expression inside the square root must be non-negative.

Let's solve these conditions step by step.

### Step 1: Ensure the argument of the logarithm is positive

We need:
\[ \frac{x-1}{x+5} > 0 \]

This inequality holds true when both the numerator and denominator are either both positive or both negative.

**Finding critical points:**
- The numerator \( x - 1 = 0 \) gives \( x = 1 \).
- The denominator \( x + 5 = 0 \) gives \( x = -5 \).

**Test intervals:**
We will test the intervals determined by the critical points \( -5 \) and \( 1 \):
- Interval 1: \( (-\infty, -5) \)
- Interval 2: \( (-5, 1) \)
- Interval 3: \( (1, \infty) \)

**Testing the intervals:**
1. For \( x < -5 \) (e.g., \( x = -6 \)):
\[ \frac{-6 - 1}{-6 + 5} = \frac{-7}{-1} = 7 > 0 \quad \text{(valid)} \]
2. For \( -5 < x < 1 \) (e.g., \( x = 0 \)):
\[ \frac{0 - 1}{0 + 5} = \frac{-1}{5} < 0 \quad \text{(invalid)} \]
3. For \( x > 1 \) (e.g., \( x = 2 \)):
\[ \frac{2 - 1}{2 + 5} = \frac{1}{7} > 0 \quad \text{(valid)} \]

**Conclusion for this step:**
The valid intervals for \( \frac{x-1}{x+5} > 0 \) are \( (-\infty, -5) \) and \( (1, \infty) \).

### Step 2: Ensure the logarithm is non-negative

Next, we need:
\[ \log_{1/2} \left( \frac{x-1}{x+5} \right) \geq 0 \]

Since the base \( \frac{1}{2} \) is less than 1, the logarithm is non-negative when:
\[ \frac{x-1}{x+5} \leq 1 \]

**Solving the inequality:**
\[ \frac{x-1}{x+5} \leq 1 \]
Subtract \( 1 \) from both sides:
\[ \frac{x-1}{x+5} - 1 \leq 0 \]
This simplifies to:
\[ \frac{x-1 - (x+5)}{x+5} \leq 0 \]
\[ \frac{-6}{x+5} \leq 0 \]

This inequality holds when \( x + 5 > 0 \) (because \( -6 \) is negative). Thus:
\[ x + 5 > 0 \implies x > -5 \]

### Step 3: Combine the results

From the two steps, we have:
1. \( x \in (-\infty, -5) \cup (1, \infty) \) from the first condition.
2. \( x > -5 \) from the second condition.

The common valid region is:
\[ x \in (1, \infty) \]

### Final Answer:
The values of \( x \) for which the function \( f(x) \) is defined are:
\[ \boxed{(1, \infty)} \]