Solve : `(1)/(x^(2) +x) le(1)/(2x^(2) + 2x+3)`

Answer

To solve the inequality

\[ \frac{1}{x^2 + x} \leq \frac{1}{2x^2 + 2x + 3} \]

we will follow these steps:

### Step 1: Rewrite the inequality
We start by rewriting the inequality:

\[ \frac{1}{x^2 + x} - \frac{1}{2x^2 + 2x + 3} \leq 0 \]

### Step 2: Find a common denominator
The common denominator of the two fractions is \((x^2 + x)(2x^2 + 2x + 3)\). Thus, we can rewrite the inequality as:

\[ \frac{(2x^2 + 2x + 3) - (x^2 + x)}{(x^2 + x)(2x^2 + 2x + 3)} \leq 0 \]

### Step 3: Simplify the numerator
Now, we simplify the numerator:

\[ 2x^2 + 2x + 3 - x^2 - x = x^2 + x + 3 \]

So the inequality becomes:

\[ \frac{x^2 + x + 3}{(x^2 + x)(2x^2 + 2x + 3)} \leq 0 \]

### Step 4: Analyze the numerator and denominator
1. **Numerator**: The quadratic \(x^2 + x + 3\) has a discriminant of \(b^2 - 4ac = 1^2 - 4(1)(3) = 1 - 12 = -11\), which is less than 0. Therefore, \(x^2 + x + 3 > 0\) for all \(x\).

2. **Denominator**: The denominator is \((x^2 + x)(2x^2 + 2x + 3)\). The term \(2x^2 + 2x + 3\) is always positive (as shown above). We need to analyze \(x^2 + x\):

- \(x^2 + x = x(x + 1)\) is zero at \(x = 0\) and \(x = -1\).
- It is positive for \(x < -1\) and \(x > 0\), and negative for \(-1 < x < 0\).

### Step 5: Set up the inequality
Since the numerator is always positive, the sign of the entire fraction is determined by the denominator:

\[ \frac{x^2 + x + 3}{(x^2 + x)(2x^2 + 2x + 3)} \leq 0 \implies (x^2 + x) \text{ must be negative.} \]

### Step 6: Solve for the intervals
The denominator \(x^2 + x < 0\) gives us the interval:

\[ -1 < x < 0 \]

### Step 7: Exclude points where the denominator is zero
At \(x = -1\) and \(x = 0\), the denominator becomes zero, so we exclude these points from our solution.

### Final Solution
Thus, the solution to the inequality is:

\[ x \in (-1, 0) \]