To solve the inequality \( \left| \frac{x^2 - 5x + 4}{x^2 - 4} \right| \leq 1 \), we will break it down into two parts based on the definition of absolute value.
### Step 1: Set up the inequalities
The inequality \( \left| A \right| \leq B \) implies that \( -B \leq A \leq B \). Here, we have:
\[
-1 \leq \frac{x^2 - 5x + 4}{x^2 - 4} \leq 1
\]
### Step 2: Solve the first inequality
We start with the first part:
\[
\frac{x^2 - 5x + 4}{x^2 - 4} \leq 1
\]
Subtract 1 from both sides:
\[
\frac{x^2 - 5x + 4}{x^2 - 4} - 1 \leq 0
\]
This can be rewritten as:
\[
\frac{x^2 - 5x + 4 - (x^2 - 4)}{x^2 - 4} \leq 0
\]
Simplifying the numerator:
\[
\frac{-5x + 8}{x^2 - 4} \leq 0
\]
Factoring the denominator:
\[
\frac{-5x + 8}{(x - 2)(x + 2)} \leq 0
\]
### Step 3: Find critical points
Set the numerator and denominator to zero to find critical points:
1. \( -5x + 8 = 0 \) gives \( x = \frac{8}{5} \)
2. \( x - 2 = 0 \) gives \( x = 2 \)
3. \( x + 2 = 0 \) gives \( x = -2 \)
### Step 4: Test intervals
We will test the sign of \( \frac{-5x + 8}{(x - 2)(x + 2)} \) in the intervals determined by the critical points: \( (-\infty, -2) \), \( (-2, \frac{8}{5}) \), \( (\frac{8}{5}, 2) \), and \( (2, \infty) \).
1. **Interval \( (-\infty, -2) \)**: Choose \( x = -3 \):
\[
\frac{-5(-3) + 8}{(-3 - 2)(-3 + 2)} = \frac{15 + 8}{(-5)(-1)} = \frac{23}{5} > 0
\]
2. **Interval \( (-2, \frac{8}{5}) \)**: Choose \( x = 0 \):
\[
\frac{-5(0) + 8}{(0 - 2)(0 + 2)} = \frac{8}{(-2)(2)} = \frac{8}{-4} < 0
\]
3. **Interval \( (\frac{8}{5}, 2) \)**: Choose \( x = 1.5 \):
\[
\frac{-5(1.5) + 8}{(1.5 - 2)(1.5 + 2)} = \frac{-7.5 + 8}{(-0.5)(3.5)} = \frac{0.5}{-1.75} < 0
\]
4. **Interval \( (2, \infty) \)**: Choose \( x = 3 \):
\[
\frac{-5(3) + 8}{(3 - 2)(3 + 2)} = \frac{-15 + 8}{(1)(5)} = \frac{-7}{5} < 0
\]
### Step 5: Combine results
The solution to the first inequality is:
\[
x \in (-2, \frac{8}{5}] \cup (2, \infty)
\]
### Step 6: Solve the second inequality
Now we solve the second part:
\[
\frac{x^2 - 5x + 4}{x^2 - 4} \geq -1
\]
Rearranging gives:
\[
\frac{x^2 - 5x + 4 + (x^2 - 4)}{x^2 - 4} \geq 0
\]
This simplifies to:
\[
\frac{2x^2 - 5x}{(x - 2)(x + 2)} \geq 0
\]
Factoring the numerator:
\[
\frac{x(2x - 5)}{(x - 2)(x + 2)} \geq 0
\]
### Step 7: Find critical points
Set the numerator and denominator to zero:
1. \( x = 0 \)
2. \( 2x - 5 = 0 \) gives \( x = \frac{5}{2} \)
3. \( x - 2 = 0 \) gives \( x = 2 \)
4. \( x + 2 = 0 \) gives \( x = -2 \)
### Step 8: Test intervals
Test the sign of \( \frac{x(2x - 5)}{(x - 2)(x + 2)} \) in the intervals determined by the critical points: \( (-\infty, -2) \), \( (-2, 0) \), \( (0, 2) \), \( (2, \frac{5}{2}) \), and \( (\frac{5}{2}, \infty) \).
1. **Interval \( (-\infty, -2) \)**: Positive
2. **Interval \( (-2, 0) \)**: Negative
3. **Interval \( (0, 2) \)**: Negative
4. **Interval \( (2, \frac{5}{2}) \)**: Positive
5. **Interval \( (\frac{5}{2}, \infty) \)**: Positive
### Step 9: Combine results
The solution to the second inequality is:
\[
x \in (-\infty, -2) \cup [0, 2) \cup \left[\frac{5}{2}, \infty\right)
\]
### Step 10: Find the intersection
Now we find the intersection of the two solutions:
1. From the first inequality: \( (-2, \frac{8}{5}] \cup (2, \infty) \)
2. From the second inequality: \( (-\infty, -2) \cup [0, 2) \cup \left[\frac{5}{2}, \infty\right) \)
The intersection gives:
\[
[0, \frac{8}{5}] \cup \left[\frac{5}{2}, \infty\right)
\]
### Final Answer
Thus, the solution to the inequality \( \left| \frac{x^2 - 5x + 4}{x^2 - 4} \right| \leq 1 \) is:
\[
[0, \frac{8}{5}] \cup \left[\frac{5}{2}, \infty\right)
\]
