If `log _( 0.6) (log _(6) ((x ^(2) +x)/(x +4))) lt 0,` then complete set of value of 'x' is:

Options

A. `(-4, -3) uu (8,oo)`
B. `(-oo, -3)uu(8,oo)`
C. `(8,oo)`
D. none of these

Answer

To solve the inequality \( \log_{0.6} \left( \log_{6} \left( \frac{x^2 + x}{x + 4} \right) \right) < 0 \), we will break it down into steps.

### Step 1: Understand the Logarithmic Inequality
The inequality \( \log_{0.6}(A) < 0 \) implies that \( A > 1 \) since the base \( 0.6 < 1 \) reverses the inequality. Therefore, we need to solve:
\[ \log_{6} \left( \frac{x^2 + x}{x + 4} \right) > 1 \]

### Step 2: Exponentiate to Remove the Logarithm
Exponentiating both sides with base 6 gives:
\[ \frac{x^2 + x}{x + 4} > 6^1 \]
This simplifies to:
\[ \frac{x^2 + x}{x + 4} > 6 \]

### Step 3: Rearranging the Inequality
To solve the inequality, we can rearrange it:
\[ x^2 + x > 6(x + 4) \]
Expanding the right side:
\[ x^2 + x > 6x + 24 \]
Rearranging gives:
\[ x^2 - 5x - 24 > 0 \]

### Step 4: Factor the Quadratic Expression
Now, we will factor the quadratic:
\[ x^2 - 5x - 24 = (x - 8)(x + 3) \]
Thus, the inequality becomes:
\[ (x - 8)(x + 3) > 0 \]

### Step 5: Determine the Critical Points
The critical points are \( x = 8 \) and \( x = -3 \). We will analyze the sign of the product in the intervals determined by these points: \( (-\infty, -3) \), \( (-3, 8) \), and \( (8, \infty) \).

### Step 6: Test the Intervals
1. **Interval \( (-\infty, -3) \)**: Choose \( x = -4 \):
\[ (-4 - 8)(-4 + 3) = (-12)(-1) > 0 \quad \text{(True)} \]
2. **Interval \( (-3, 8) \)**: Choose \( x = 0 \):
\[ (0 - 8)(0 + 3) = (-8)(3) < 0 \quad \text{(False)} \]
3. **Interval \( (8, \infty) \)**: Choose \( x = 9 \):
\[ (9 - 8)(9 + 3) = (1)(12) > 0 \quad \text{(True)} \]

### Step 7: Combine the Results
The solution to the inequality \( (x - 8)(x + 3) > 0 \) is:
\[ x \in (-\infty, -3) \cup (8, \infty) \]

### Step 8: Consider the Domain of the Logarithm
Next, we must ensure that the argument of the logarithm \( \frac{x^2 + x}{x + 4} \) is positive:
\[ \frac{x^2 + x}{x + 4} > 0 \]
This requires \( x^2 + x > 0 \) and \( x + 4 > 0 \).

### Step 9: Solve for \( x^2 + x > 0 \)
Factoring gives:
\[ x(x + 1) > 0 \]
The critical points are \( x = 0 \) and \( x = -1 \). Testing intervals:
- For \( (-\infty, -1) \): True
- For \( (-1, 0) \): False
- For \( (0, \infty) \): True

Thus, \( x \in (-\infty, -1) \cup (0, \infty) \).

### Step 10: Combine Conditions
Now we combine the two sets of solutions:
1. From the quadratic inequality: \( x \in (-\infty, -3) \cup (8, \infty) \)
2. From the logarithm condition: \( x \in (-\infty, -1) \cup (0, \infty) \)

The common solution is:
\[ x \in (-\infty, -3) \cup (8, \infty) \]

### Final Answer
The complete set of values of \( x \) satisfying the original inequality is:
\[ \boxed{(-\infty, -3) \cup (8, \infty)} \]