Solve the following systems of inequalities for all `x in R`
Solve `(|x|-1)(|x|-2)ge0`,`x in R`,`x` is not equal to `pm2`

Answer

To solve the inequality \((|x|-1)(|x|-2) \geq 0\) with the condition that \(x \neq \pm 2\), we will follow these steps:

### Step 1: Identify Critical Points
The expression \((|x|-1)(|x|-2)\) will be zero at the points where \(|x| = 1\) and \(|x| = 2\). Thus, the critical points are:
- \(|x| = 1\) gives us \(x = 1\) and \(x = -1\).
- \(|x| = 2\) gives us \(x = 2\) and \(x = -2\).

### Step 2: Determine Intervals
The critical points divide the real number line into the following intervals:
1. \( (-\infty, -2) \)
2. \( (-2, -1) \)
3. \( (-1, 1) \)
4. \( (1, 2) \)
5. \( (2, \infty) \)

### Step 3: Test Each Interval
We will test a point from each interval to determine where the expression \((|x|-1)(|x|-2)\) is non-negative.

1. **Interval \( (-\infty, -2) \)**: Choose \(x = -3\)
\[ (|-3|-1)(|-3|-2) = (3-1)(3-2) = 2 \cdot 1 = 2 \quad (\text{positive}) \]

2. **Interval \( (-2, -1) \)**: Choose \(x = -1.5\)
\[ (|-1.5|-1)(|-1.5|-2) = (1.5-1)(1.5-2) = 0.5 \cdot (-0.5) = -0.25 \quad (\text{negative}) \]

3. **Interval \( (-1, 1) \)**: Choose \(x = 0\)
\[ (|0|-1)(|0|-2) = (-1)(-2) = 2 \quad (\text{positive}) \]

4. **Interval \( (1, 2) \)**: Choose \(x = 1.5\)
\[ (|1.5|-1)(|1.5|-2) = (1.5-1)(1.5-2) = 0.5 \cdot (-0.5) = -0.25 \quad (\text{negative}) \]

5. **Interval \( (2, \infty) \)**: Choose \(x = 3\)
\[ (|3|-1)(|3|-2) = (3-1)(3-2) = 2 \cdot 1 = 2 \quad (\text{positive}) \]

### Step 4: Compile Results
From our tests, we find:
- The expression is non-negative in the intervals \( (-\infty, -2) \), \( (-1, 1) \), and \( (2, \infty) \).
- At the critical points \(x = -1\) and \(x = 1\), the expression equals zero.
- However, \(x = 2\) and \(x = -2\) are not included in our solution set.

### Step 5: Write the Final Solution
Combining the intervals where the expression is non-negative and considering the restrictions, we have:
\[ x \in (-\infty, -2) \cup [-1, 1] \cup (2, \infty) \]