Solve the following inequalities:
`((x-1)(x-2))/((2x-5)(x+4))lt0`

Answer

To solve the inequality \(\frac{(x-1)(x-2)}{(2x-5)(x+4)} < 0\), 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:**
- \(x - 1 = 0 \Rightarrow x = 1\)
- \(x - 2 = 0 \Rightarrow x = 2\)

2. **Denominator:**
- \(2x - 5 = 0 \Rightarrow 2x = 5 \Rightarrow x = \frac{5}{2}\)
- \(x + 4 = 0 \Rightarrow x = -4\)

So, the critical points are \(x = -4, 1, 2, \frac{5}{2}\).

### Step 2: Determine the intervals
The critical points divide the number line into the following intervals:
1. \((-∞, -4)\)
2. \((-4, 1)\)
3. \((1, 2)\)
4. \((2, \frac{5}{2})\)
5. \((\frac{5}{2}, ∞)\)

### Step 3: Test each interval
We will select a test point from each interval to determine the sign of the expression \(\frac{(x-1)(x-2)}{(2x-5)(x+4)}\).

1. **Interval \((-∞, -4)\)**:
- Test point: \(x = -5\)
- \(\frac{(-5-1)(-5-2)}{(2(-5)-5)(-5+4)} = \frac{(-6)(-7)}{(-10-5)(-1)} = \frac{42}{(-15)(-1)} > 0\)

2. **Interval \((-4, 1)\)**:
- Test point: \(x = 0\)
- \(\frac{(0-1)(0-2)}{(2(0)-5)(0+4)} = \frac{(-1)(-2)}{(-5)(4)} = \frac{2}{-20} < 0\)

3. **Interval \((1, 2)\)**:
- Test point: \(x = 1.5\)
- \(\frac{(1.5-1)(1.5-2)}{(2(1.5)-5)(1.5+4)} = \frac{(0.5)(-0.5)}{(3-5)(5.5)} = \frac{-0.25}{-2(5.5)} > 0\)

4. **Interval \((2, \frac{5}{2})\)**:
- Test point: \(x = 2.2\)
- \(\frac{(2.2-1)(2.2-2)}{(2(2.2)-5)(2.2+4)} = \frac{(1.2)(0.2)}{(4.4-5)(6.2)} = \frac{0.24}{-0.6(6.2)} < 0\)

5. **Interval \((\frac{5}{2}, ∞)\)**:
- Test point: \(x = 3\)
- \(\frac{(3-1)(3-2)}{(2(3)-5)(3+4)} = \frac{(2)(1)}{(6-5)(7)} = \frac{2}{1(7)} > 0\)

### Step 4: Compile the results
From our tests, we find that the expression is negative in the intervals:
- \((-4, 1)\)
- \((2, \frac{5}{2})\)

### Step 5: Write the solution
Since we are looking for where the expression is less than zero, we can write the solution as:
\[ x \in (-4, 1) \cup (2, \frac{5}{2}) \]