The set of values of `x` satisfying `|(x^(2)-5x+4)/(x^(2)-4)| le 1` is

Options

A. `[0,(8)/(5)]uu[5/2,oo)`
B. `[(8)/(5),oo]`
C. `[-oo,-2]`
D. `R`

Explanation

To solve the inequality \( \left| \frac{x^2 - 5x + 4}{x^2 - 4} \right| \leq 1 \), we will break it down into two separate inequalities:

1. \( \frac{x^2 - 5x + 4}{x^2 - 4} \leq 1 \)
2. \( \frac{x^2 - 5x + 4}{x^2 - 4} \geq -1 \)

### Step 1: Solve the first inequality \( \frac{x^2 - 5x + 4}{x^2 - 4} \leq 1 \)

Rearranging gives:
\[ \frac{x^2 - 5x + 4}{x^2 - 4} - 1 \leq 0 \]
This simplifies to:
\[ \frac{x^2 - 5x + 4 - (x^2 - 4)}{x^2 - 4} \leq 0 \]
\[ \frac{-5x + 8}{x^2 - 4} \leq 0 \]

### Step 2: Factor the numerator and denominator

The numerator \( -5x + 8 \) can be factored as:
\[ -5(x - \frac{8}{5}) \]
The denominator \( x^2 - 4 \) factors to:
\[ (x - 2)(x + 2) \]

Thus, we have:
\[ \frac{-5(x - \frac{8}{5})}{(x - 2)(x + 2)} \leq 0 \]

### Step 3: Identify critical points

The critical points are:
- \( x = \frac{8}{5} \)
- \( x = 2 \)
- \( x = -2 \)

### Step 4: Test intervals on a number line

We will test 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 \( x = -3 \)
\[ \frac{-5(-3 - \frac{8}{5})}{(-3 - 2)(-3 + 2)} = \frac{-5(-3 - 1.6)}{(-5)(-1)} > 0 \]

2. **Interval \( (-2, \frac{8}{5}) \)**: Choose \( x = 0 \)
\[ \frac{-5(0 - \frac{8}{5})}{(0 - 2)(0 + 2)} = \frac{-5(-\frac{8}{5})}{(-2)(2)} = \frac{8}{4} > 0 \]

3. **Interval \( (\frac{8}{5}, 2) \)**: Choose \( x = 1.5 \)
\[ \frac{-5(1.5 - \frac{8}{5})}{(1.5 - 2)(1.5 + 2)} < 0 \]

4. **Interval \( (2, \infty) \)**: Choose \( x = 3 \)
\[ \frac{-5(3 - \frac{8}{5})}{(3 - 2)(3 + 2)} > 0 \]

### Step 5: Combine results

From the tests, we find that the inequality holds in the interval \( \left( \frac{8}{5}, 2 \right) \).

### Step 6: Solve the second inequality \( \frac{x^2 - 5x + 4}{x^2 - 4} \geq -1 \)

Rearranging gives:
\[ \frac{x^2 - 5x + 4}{x^2 - 4} + 1 \geq 0 \]
This simplifies to:
\[ \frac{x^2 - 5x + 4 + (x^2 - 4)}{x^2 - 4} \geq 0 \]
\[ \frac{2x^2 - 5x}{x^2 - 4} \geq 0 \]

### Step 7: Factor the numerator and denominator

The numerator \( 2x^2 - 5x \) factors to:
\[ x(2x - 5) \]
Thus, we have:
\[ \frac{x(2x - 5)}{(x - 2)(x + 2)} \geq 0 \]

### Step 8: Identify critical points

The critical points are:
- \( x = 0 \)
- \( x = \frac{5}{2} \)
- \( x = 2 \)
- \( x = -2 \)

### Step 9: Test intervals on a number line

We will test the intervals determined by the critical points: \( (-\infty, -2) \), \( (-2, 0) \), \( (0, 2) \), \( (2, \frac{5}{2}) \), and \( (\frac{5}{2}, \infty) \).

1. **Interval \( (-\infty, -2) \)**: Choose \( x = -3 \)
\[ \frac{-3(2(-3) - 5)}{(-3 - 2)(-3 + 2)} > 0 \]

2. **Interval \( (-2, 0) \)**: Choose \( x = -1 \)
\[ \frac{-1(2(-1) - 5)}{(-1 - 2)(-1 + 2)} > 0 \]

3. **Interval \( (0, 2) \)**: Choose \( x = 1 \)
\[ \frac{1(2(1) - 5)}{(1 - 2)(1 + 2)} < 0 \]

4. **Interval \( (2, \frac{5}{2}) \)**: Choose \( x = 2.2 \)
\[ \frac{2.2(2(2.2) - 5)}{(2.2 - 2)(2.2 + 2)} > 0 \]

5. **Interval \( (\frac{5}{2}, \infty) \)**: Choose \( x = 3 \)
\[ \frac{3(2(3) - 5)}{(3 - 2)(3 + 2)} > 0 \]

### Step 10: Combine results

From the tests, we find that the inequality holds in the intervals \( (-\infty, -2) \), \( (-2, 0) \), \( (2, \frac{5}{2}) \), and \( (\frac{5}{2}, \infty) \).

### Step 11: Find the common intervals

The common intervals from both inequalities are:
- \( (0, \frac{8}{5}) \)
- \( (2, \frac{5}{2}) \)

### Final Answer

Thus, the set of values of \( x \) satisfying the original inequality is:
\[ \boxed{(0, \frac{8}{5}) \cup (2, \frac{5}{2})} \]