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
The critical points occur where the numerator and denominator are equal to zero.
- **Numerator**:
- \(x - 1 = 0 \Rightarrow x = 1\)
- \(x - 2 = 0 \Rightarrow x = 2\)
- **Denominator**:
- \(2x - 5 = 0 \Rightarrow x = \frac{5}{2}\)
- \(x + 4 = 0 \Rightarrow x = -4\)
The critical points are \(x = -4, 1, 2, \frac{5}{2}\).
### Step 2: Plot the critical points on a number line
We will plot the critical points on a number line to divide it into intervals:
```
---|----|----|----|----|----|----|----
-4 1 2 2.5
```
### Step 3: Test the intervals
We will test the sign of the expression in each interval created by these critical points. The intervals are:
1. \( (-\infty, -4) \)
2. \( (-4, 1) \)
3. \( (1, 2) \)
4. \( (2, \frac{5}{2}) \)
5. \( (\frac{5}{2}, \infty) \)
**Interval 1: \( (-\infty, -4) \)**
- Choose \(x = -5\):
\[
\frac{(-5-1)(-5-2)}{(2(-5)-5)(-5+4)} = \frac{(-6)(-7)}{(-15)(-1)} = \frac{42}{15} > 0
\]
**Interval 2: \( (-4, 1) \)**
- Choose \(x = 0\):
\[
\frac{(0-1)(0-2)}{(2(0)-5)(0+4)} = \frac{(-1)(-2)}{(-5)(4)} = \frac{2}{-20} < 0
\]
**Interval 3: \( (1, 2) \)**
- Choose \(x = 1.5\):
\[
\frac{(1.5-1)(1.5-2)}{(2(1.5)-5)(1.5+4)} = \frac{(0.5)(-0.5)}{(3-5)(5.5)} = \frac{-0.25}{-11} > 0
\]
**Interval 4: \( (2, \frac{5}{2}) \)**
- Choose \(x = 2.2\):
\[
\frac{(2.2-1)(2.2-2)}{(2(2.2)-5)(2.2+4)} = \frac{(1.2)(0.2)}{(4.4-5)(6.2)} = \frac{0.24}{-0.6} < 0
\]
**Interval 5: \( (\frac{5}{2}, \infty) \)**
- Choose \(x = 3\):
\[
\frac{(3-1)(3-2)}{(2(3)-5)(3+4)} = \frac{(2)(1)}{(6-5)(7)} = \frac{2}{7} > 0
\]
### Step 4: Compile the results
From our tests, we find:
- The expression is negative in the intervals \( (-4, 1) \) and \( (2, \frac{5}{2}) \).
### Step 5: Write the solution
Since the inequality is strict (\(< 0\)), we do not include the endpoints where the expression equals zero or is undefined. Therefore, the solution is:
\[
x \in (-4, 1) \cup (2, \frac{5}{2})
\]
