The solution of the inequality
`x log_((1)/(10)) (x^(2) + x + 1) gt 0 `
is given by

Options

A. `-oo lt x lt 2`
B. `-oo lt x lt - 1`
C. `- 3 lt x lt 1`
D. `1 lt x lt 3 `

Answer

To solve the inequality \( x \log_{(1/10)}(x^2 + x + 1) > 0 \), we will follow these steps:

### Step 1: Analyze the logarithmic function
The logarithmic function \( \log_{(1/10)}(x^2 + x + 1) \) is defined only when its argument \( x^2 + x + 1 > 0 \).

### Step 2: Determine when \( x^2 + x + 1 > 0 \)
To find the roots of the quadratic equation \( x^2 + x + 1 = 0 \), we calculate the discriminant:
\[ D = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]
Since the discriminant is negative, the quadratic has no real roots and is always positive (as the coefficient of \( x^2 \) is positive). Therefore, \( x^2 + x + 1 > 0 \) for all \( x \).

### Step 3: Rewrite the logarithm
We can rewrite the logarithm using the change of base formula:
\[ \log_{(1/10)}(x^2 + x + 1) = \frac{\log_{10}(x^2 + x + 1)}{\log_{10}(1/10)} = -\log_{10}(x^2 + x + 1) \]
Thus, the inequality becomes:
\[ x \cdot (-\log_{10}(x^2 + x + 1)) > 0 \]
This simplifies to:
\[ -x \log_{10}(x^2 + x + 1) > 0 \]

### Step 4: Analyze the sign of the product
The product \( -x \log_{10}(x^2 + x + 1) > 0 \) implies that \( -x \) and \( \log_{10}(x^2 + x + 1) \) must have opposite signs.

### Step 5: Case analysis
1. **Case 1:** \( x > 0 \)
- Here, \( -x < 0 \). For the product to be positive, \( \log_{10}(x^2 + x + 1) < 0 \).
- This means \( x^2 + x + 1 < 1 \), which simplifies to \( x^2 + x < 0 \).
- Factoring gives \( x(x + 1) < 0 \), which has roots at \( x = 0 \) and \( x = -1 \). The solution is \( -1 < x < 0 \). However, this contradicts \( x > 0 \), so there are no solutions from this case.

2. **Case 2:** \( x < 0 \)
- Here, \( -x > 0 \). For the product to be positive, \( \log_{10}(x^2 + x + 1) > 0 \).
- This means \( x^2 + x + 1 > 1 \), which simplifies to \( x^2 + x > 0 \).
- Factoring gives \( x(x + 1) > 0 \), which has roots at \( x = 0 \) and \( x = -1 \). The solution is \( x < -1 \) or \( x > 0 \). Since we are in the case where \( x < 0 \), we only take \( x < -1 \).

### Final Solution
Combining the results from both cases, we find that the solution to the inequality is:
\[ x \in (-\infty, -1) \]