Solve `((x-1)(x-2))/((x-3)(x-4))ge0`

Answer

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

### Step 1: Identify the points where the expression is zero or undefined
The expression \(\frac{(x-1)(x-2)}{(x-3)(x-4)}\) is zero when the numerator is zero. This occurs at:
- \(x - 1 = 0 \Rightarrow x = 1\)
- \(x - 2 = 0 \Rightarrow x = 2\)

The expression is undefined when the denominator is zero. This occurs at:
- \(x - 3 = 0 \Rightarrow x = 3\)
- \(x - 4 = 0 \Rightarrow x = 4\)

### Step 2: Create a number line with critical points
The critical points we found are \(1, 2, 3, 4\). We will use these points to divide the number line into intervals:
- \((-∞, 1)\)
- \((1, 2)\)
- \((2, 3)\)
- \((3, 4)\)
- \((4, ∞)\)

### Step 3: Test each interval
We will test a point from each interval to determine the sign of the expression in that interval.

1. **Interval \((-∞, 1)\)**: Choose \(x = 0\)
\[ \frac{(0-1)(0-2)}{(0-3)(0-4)} = \frac{(-1)(-2)}{(-3)(-4)} = \frac{2}{12} = \frac{1}{6} > 0 \]
(Positive)

2. **Interval \((1, 2)\)**: Choose \(x = 1.5\)
\[ \frac{(1.5-1)(1.5-2)}{(1.5-3)(1.5-4)} = \frac{(0.5)(-0.5)}{(-1.5)(-2.5)} = \frac{-0.25}{3.75} < 0 \]
(Negative)

3. **Interval \((2, 3)\)**: Choose \(x = 2.5\)
\[ \frac{(2.5-1)(2.5-2)}{(2.5-3)(2.5-4)} = \frac{(1.5)(0.5)}{(-0.5)(-1.5)} = \frac{0.75}{0.75} = 1 > 0 \]
(Positive)

4. **Interval \((3, 4)\)**: Choose \(x = 3.5\)
\[ \frac{(3.5-1)(3.5-2)}{(3.5-3)(3.5-4)} = \frac{(2.5)(1.5)}{(0.5)(-0.5)} = \frac{3.75}{-0.25} < 0 \]
(Negative)

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

### Step 4: Determine the intervals where the expression is greater than or equal to zero
From our tests, we have:
- Positive in \((-∞, 1)\)
- Negative in \((1, 2)\)
- Positive in \((2, 3)\)
- Negative in \((3, 4)\)
- Positive in \((4, ∞)\)

### Step 5: Include the points where the expression is zero
The expression is equal to zero at \(x = 1\) and \(x = 2\). It is undefined at \(x = 3\) and \(x = 4\).

### Final Solution
Thus, the solution set is:
\[ x \in (-\infty, 1] \cup [2, 3) \cup (4, \infty) \]