`(2x-1)/(2x^(3)+3x^(2)+x)gt 0.`

Answer

To solve the inequality \(\frac{2x-1}{2x^3 + 3x^2 + x} > 0\), we will follow these steps:

### Step 1: Factor the denominator
First, we need to factor the denominator \(2x^3 + 3x^2 + x\).

1. Take \(x\) common from the denominator:
\[ 2x^3 + 3x^2 + x = x(2x^2 + 3x + 1) \]

2. Now, we need to factor the quadratic \(2x^2 + 3x + 1\). We can find its factors by using the quadratic formula or by inspection:
\[ 2x^2 + 3x + 1 = (x + 1)(2x + 1) \]

Thus, the complete factorization of the denominator is:
\[ 2x^3 + 3x^2 + x = x(x + 1)(2x + 1) \]

### Step 2: Identify critical points
Next, we need to find the critical points where the expression is either zero or undefined. This occurs at the points where the numerator and denominator are zero.

1. Set the numerator \(2x - 1 = 0\):
\[ 2x - 1 = 0 \implies x = \frac{1}{2} \]

2. Set the denominator \(x(x + 1)(2x + 1) = 0\):
- \(x = 0\)
- \(x + 1 = 0 \implies x = -1\)
- \(2x + 1 = 0 \implies x = -\frac{1}{2}\)

The critical points are:
\[ x = -1, \quad x = -\frac{1}{2}, \quad x = 0, \quad x = \frac{1}{2} \]

### Step 3: Test intervals
Now, we will test the sign of the expression in the intervals determined by the critical points. The intervals are:
1. \((- \infty, -1)\)
2. \((-1, -\frac{1}{2})\)
3. \((- \frac{1}{2}, 0)\)
4. \((0, \frac{1}{2})\)
5. \((\frac{1}{2}, +\infty)\)

We will use test points from each interval to determine the sign of the expression.

1. **Interval \((- \infty, -1)\)**: Test \(x = -2\)
\[ \frac{2(-2) - 1}{2(-2)^3 + 3(-2)^2 + (-2)} = \frac{-4 - 1}{-16 + 12 - 2} = \frac{-5}{-6} > 0 \]

2. **Interval \((-1, -\frac{1}{2})\)**: Test \(x = -0.75\)
\[ \frac{2(-0.75) - 1}{2(-0.75)^3 + 3(-0.75)^2 + (-0.75)} = \frac{-1.5 - 1}{-0.84375 + 1.6875 - 0.75} = \frac{-2.5}{-0.90625} > 0 \]

3. **Interval \((- \frac{1}{2}, 0)\)**: Test \(x = -0.25\)
\[ \frac{2(-0.25) - 1}{2(-0.25)^3 + 3(-0.25)^2 + (-0.25)} = \frac{-0.5 - 1}{-0.125 + 0.1875 - 0.25} = \frac{-1.5}{-0.1875} > 0 \]

4. **Interval \((0, \frac{1}{2})\)**: Test \(x = 0.25\)
\[ \frac{2(0.25) - 1}{2(0.25)^3 + 3(0.25)^2 + (0.25)} = \frac{0.5 - 1}{0.125 + 0.1875 + 0.25} = \frac{-0.5}{0.5625} < 0 \]

5. **Interval \((\frac{1}{2}, +\infty)\)**: Test \(x = 1\)
\[ \frac{2(1) - 1}{2(1)^3 + 3(1)^2 + (1)} = \frac{2 - 1}{2 + 3 + 1} = \frac{1}{6} > 0 \]

### Step 4: Combine results
The expression is positive in the intervals:
- \((- \infty, -1)\)
- \((-1, -\frac{1}{2})\)
- \((- \frac{1}{2}, 0)\)
- \((\frac{1}{2}, +\infty)\)

### Final Answer
Thus, the solution to the inequality \(\frac{2x-1}{2x^3 + 3x^2 + x} > 0\) is:
\[ x \in (-\infty, -1) \cup (-1, -\frac{1}{2}) \cup (-\frac{1}{2}, 0) \cup (\frac{1}{2}, +\infty) \]