Solve the in equality `log_(1/4) (2-x)>log_(1/4) (2/(x+1))`.

Answer

To solve the inequality \( \log_{1/4} (2 - x) > \log_{1/4} \left(\frac{2}{x + 1}\right) \), we will follow these steps:

### Step 1: Define the domain of the logarithmic functions
The arguments of the logarithms must be positive.

1. For \( \log_{1/4} (2 - x) \):
\[ 2 - x > 0 \implies x < 2 \]

2. For \( \log_{1/4} \left(\frac{2}{x + 1}\right) \):
\[ \frac{2}{x + 1} > 0 \implies x + 1 > 0 \implies x > -1 \]

### Step 2: Combine the domain conditions
From the above, we have:
- \( x < 2 \)
- \( x > -1 \)

Thus, the combined domain is:
\[ -1 < x < 2 \]

### Step 3: Rewrite the inequality
Using the property of logarithms, we can rewrite the inequality:
\[ \log_{1/4} (2 - x) - \log_{1/4} \left(\frac{2}{x + 1}\right) > 0 \]
This can be expressed as:
\[ \log_{1/4} \left(\frac{2 - x}{\frac{2}{x + 1}}\right) > 0 \]
which simplifies to:
\[ \log_{1/4} \left((2 - x)(x + 1) / 2\right) > 0 \]

### Step 4: Convert the logarithmic inequality
Since the base \( \frac{1}{4} < 1 \), we can reverse the inequality:
\[ \frac{(2 - x)(x + 1)}{2} < 1 \]

### Step 5: Solve the inequality
Multiply both sides by 2 (which is positive):
\[ (2 - x)(x + 1) < 2 \]
Expanding the left side:
\[ 2 - x^2 + x < 2 \]
Subtracting 2 from both sides gives:
\[ -x^2 + x < 0 \]
Rearranging:
\[ x^2 - x > 0 \]
Factoring:
\[ x(x - 1) > 0 \]

### Step 6: Find the critical points and test intervals
The critical points are \( x = 0 \) and \( x = 1 \). We will test the intervals:
1. \( (-\infty, 0) \)
2. \( (0, 1) \)
3. \( (1, \infty) \)

- For \( x < 0 \) (e.g., \( x = -1 \)): \( (-1)(-2) > 0 \) (True)
- For \( 0 < x < 1 \) (e.g., \( x = 0.5 \)): \( (0.5)(-0.5) < 0 \) (False)
- For \( x > 1 \) (e.g., \( x = 2 \)): \( (2)(1) > 0 \) (True)

### Step 7: Combine with the domain
From the domain \( -1 < x < 2 \) and the intervals where the inequality holds:
- Valid intervals from the inequality: \( (-\infty, 0) \) and \( (1, \infty) \)
- Combined with the domain:
- From \( (-1, 0) \)
- From \( (1, 2) \)

Thus, the solution set is:
\[ x \in (-1, 0) \cup (1, 2) \]

### Final Answer:
\[ \boxed{(-1, 0) \cup (1, 2)} \]