Solve the inequality `[x]^2-3[x]+2lt=0.`

Answer

To solve the inequality \([x]^2 - 3[x] + 2 \leq 0\), where \([x]\) denotes the greatest integer function (also known as the floor function), we will follow these steps:

### Step 1: Substitute \([x]\) with a variable
Let \(n = [x]\). The inequality then becomes:
\[ n^2 - 3n + 2 \leq 0 \]

### Step 2: Factor the quadratic expression
We can factor the quadratic expression:
\[ n^2 - 3n + 2 = (n - 1)(n - 2) \]
Thus, the inequality can be rewritten as:
\[ (n - 1)(n - 2) \leq 0 \]

### Step 3: Determine the critical points
The critical points of the inequality are \(n = 1\) and \(n = 2\). These points divide the number line into intervals. We will test the sign of the expression in each interval:
- Interval 1: \(n < 1\)
- Interval 2: \(1 \leq n \leq 2\)
- Interval 3: \(n > 2\)

### Step 4: Test the intervals
1. **For \(n < 1\)** (e.g., \(n = 0\)):
\((0 - 1)(0 - 2) = (-1)(-2) = 2 > 0\) (not a solution)

2. **For \(1 \leq n \leq 2\)** (e.g., \(n = 1.5\)):
\((1.5 - 1)(1.5 - 2) = (0.5)(-0.5) = -0.25 \leq 0\) (solution)

3. **For \(n > 2\)** (e.g., \(n = 3\)):
\((3 - 1)(3 - 2) = (2)(1) = 2 > 0\) (not a solution)

### Step 5: Combine the results
The solution to the inequality is:
\[ 1 \leq n \leq 2 \]

### Step 6: Translate back to \(x\)
Since \(n = [x]\), we have:
- If \(n = 1\), then \(1 \leq x < 2\)
- If \(n = 2\), then \(2 \leq x < 3\)

Thus, combining these intervals, we get:
\[ 1 \leq x < 3 \]

### Final Answer
The solution to the inequality is:
\[ x \in [1, 3) \]