Solve `(x(3-4x)(x+1))/(2x-5)lt 0 `

Explanation

To solve the inequality \(\frac{x(3-4x)(x+1)}{2x-5} < 0\), we will follow these steps:

### Step 1: Identify the critical points
We need to find the values of \(x\) that make the numerator and denominator equal to zero.

1. **Numerator:**
- \(x = 0\)
- \(3 - 4x = 0 \Rightarrow 4x = 3 \Rightarrow x = \frac{3}{4}\)
- \(x + 1 = 0 \Rightarrow x = -1\)

2. **Denominator:**
- \(2x - 5 = 0 \Rightarrow 2x = 5 \Rightarrow x = \frac{5}{2}\)

Thus, the critical points are \(x = -1\), \(x = 0\), \(x = \frac{3}{4}\), and \(x = \frac{5}{2}\).

### Step 2: Create a number line
We will plot the critical points on a number line:

-1, 0, \(\frac{3}{4}\), \(\frac{5}{2}\)

### Step 3: Test intervals
We will test the sign of the expression in each interval created by these critical points:

1. **Interval \((-∞, -1)\)**: Choose \(x = -2\)
\[ \frac{-2(3 - 4(-2))(-2 + 1)}{2(-2) - 5} = \frac{-2(3 + 8)(-1)}{-4 - 5} = \frac{-2 \cdot 11 \cdot -1}{-9} > 0 \]

2. **Interval \((-1, 0)\)**: Choose \(x = -0.5\)
\[ \frac{-0.5(3 - 4(-0.5))(-0.5 + 1)}{2(-0.5) - 5} = \frac{-0.5(3 + 2)(0.5)}{-1 - 5} = \frac{-0.5 \cdot 5 \cdot 0.5}{-6} > 0 \]

3. **Interval \((0, \frac{3}{4})\)**: Choose \(x = 0.5\)
\[ \frac{0.5(3 - 4(0.5))(0.5 + 1)}{2(0.5) - 5} = \frac{0.5(3 - 2)(1.5)}{1 - 5} = \frac{0.5 \cdot 1 \cdot 1.5}{-4} < 0 \]

4. **Interval \((\frac{3}{4}, \frac{5}{2})\)**: Choose \(x = 1\)
\[ \frac{1(3 - 4(1))(1 + 1)}{2(1) - 5} = \frac{1(3 - 4)(2)}{2 - 5} = \frac{1 \cdot -1 \cdot 2}{-3} > 0 \]

5. **Interval \((\frac{5}{2}, ∞)\)**: Choose \(x = 3\)
\[ \frac{3(3 - 4(3))(3 + 1)}{2(3) - 5} = \frac{3(3 - 12)(4)}{6 - 5} = \frac{3 \cdot -9 \cdot 4}{1} < 0 \]

### Step 4: Determine the solution set
The expression is negative in the intervals:
- \((0, \frac{3}{4})\)
- \((\frac{5}{2}, ∞)\)

### Step 5: Combine the intervals
The solution to the inequality \(\frac{x(3-4x)(x+1)}{2x-5} < 0\) is:
\[ x \in (0, \frac{3}{4}) \cup (\frac{5}{2}, \infty) \]

### Final Answer
\[ x \in (0, \frac{3}{4}) \cup (\frac{5}{2}, \infty) \]