The solution set of inequality
`((e^(x)-1)(2x-3)(x^(2)+x+2))/((sinx-2)(x+1)x) le 0`

Options

A. `[(3)/(2),oo)`
B. `(-oo,-1) uu [(3)/(2),oo)`
C. `(-1,0) uu [(3)/(2),oo)`
D. `R-{0,-1}`

Explanation

To solve the inequality

\[ \frac{(e^{x}-1)(2x-3)(x^{2}+x+2)}{(\sin x - 2)(x+1)x} \leq 0, \]

we will analyze each component of the inequality step by step.

### Step 1: Analyze the components of the inequality

1. **Numerator**:
- \(e^x - 1\): This is zero when \(x = 0\) and positive for \(x > 0\).
- \(2x - 3\): This is zero when \(x = \frac{3}{2}\) and positive for \(x > \frac{3}{2}\).
- \(x^2 + x + 2\): The discriminant \(D = b^2 - 4ac = 1 - 8 = -7\) is negative, indicating that this quadratic is always positive.

2. **Denominator**:
- \(\sin x - 2\): The sine function oscillates between -1 and 1, so \(\sin x - 2\) is always negative.
- \(x + 1\): This is zero when \(x = -1\) and positive for \(x > -1\).
- \(x\): This is zero when \(x = 0\) and positive for \(x > 0\).

### Step 2: Identify critical points

The critical points from the numerator and denominator are:
- From \(e^x - 1\): \(x = 0\)
- From \(2x - 3\): \(x = \frac{3}{2}\)
- From \(x + 1\): \(x = -1\)
- From \(x\): \(x = 0\) (already counted)

### Step 3: Create a number line

We will plot the critical points on a number line:
- Critical points: \(-1, 0, \frac{3}{2}\)

### Step 4: Test intervals

We will test the sign of the expression in each interval determined by the critical points:
1. **Interval \((- \infty, -1)\)**: Choose \(x = -2\)
- Numerator: Positive (since \(e^{-2} - 1 < 0\), \(2(-2) - 3 < 0\), \(x^2 + x + 2 > 0\))
- Denominator: Negative (since \(\sin(-2) - 2 < 0\), \(-2 + 1 < 0\), \(-2 < 0\))
- Overall: Positive

2. **Interval \((-1, 0)\)**: Choose \(x = -0.5\)
- Numerator: Positive
- Denominator: Negative
- Overall: Negative

3. **Interval \((0, \frac{3}{2})\)**: Choose \(x = 1\)
- Numerator: Positive
- Denominator: Negative
- Overall: Negative

4. **Interval \((\frac{3}{2}, \infty)\)**: Choose \(x = 2\)
- Numerator: Positive
- Denominator: Positive
- Overall: Positive

### Step 5: Determine the solution set

From the sign analysis:
- The expression is less than or equal to zero in the intervals \((-1, 0)\) and \((0, \frac{3}{2})\).

### Step 6: Include endpoints

- At \(x = -1\), the expression is undefined.
- At \(x = 0\), the expression is zero (included).
- At \(x = \frac{3}{2}\), the expression is zero (included).

### Final Solution

Thus, the solution set is:

\[ x \in (-1, 0] \cup [\frac{3}{2}, \infty) \]