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

Options

A. `[0,(8)/(5)]`
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 cases based on the definition of absolute value.

### Step 1: Rewrite the Inequality
We can express the inequality as 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 2: Solve the First Inequality
Starting with the first inequality:
\[ \frac{x^2 - 5x + 4}{x^2 - 4} \leq 1 \]
Subtracting 1 from both sides gives:
\[ \frac{x^2 - 5x + 4 - (x^2 - 4)}{x^2 - 4} \leq 0 \]
This simplifies to:
\[ \frac{-5x + 8}{x^2 - 4} \leq 0 \]
Factoring the denominator gives:
\[ \frac{-5x + 8}{(x - 2)(x + 2)} \leq 0 \]

### Step 3: Find Critical Points
The critical points occur when the numerator or denominator equals zero:
1. \( -5x + 8 = 0 \Rightarrow x = \frac{8}{5} \)
2. \( x - 2 = 0 \Rightarrow x = 2 \)
3. \( x + 2 = 0 \Rightarrow x = -2 \)

### Step 4: Test Intervals
We will test the intervals determined by the critical points: \((-∞, -2)\), \((-2, \frac{8}{5})\), \((\frac{8}{5}, 2)\), and \((2, ∞)\).

- For \(x < -2\) (e.g., \(x = -3\)):
\[ \frac{-5(-3) + 8}{(-3 - 2)(-3 + 2)} = \frac{15 + 8}{(-5)(-1)} = \frac{23}{5} > 0 \]
- For \(-2 < x < \frac{8}{5}\) (e.g., \(x = 0\)):
\[ \frac{-5(0) + 8}{(0 - 2)(0 + 2)} = \frac{8}{(-2)(2)} = \frac{8}{-4} < 0 \]
- For \(\frac{8}{5} < x < 2\) (e.g., \(x = 1.6\)):
\[ \frac{-5(1.6) + 8}{(1.6 - 2)(1.6 + 2)} = \frac{-8 + 8}{(-0.4)(3.6)} = \frac{0}{-1.44} = 0 \]
- For \(x > 2\) (e.g., \(x = 3\)):
\[ \frac{-5(3) + 8}{(3 - 2)(3 + 2)} = \frac{-15 + 8}{(1)(5)} = \frac{-7}{5} < 0 \]

### Step 5: Combine Results
From the tests, we find:
- The inequality is satisfied in the intervals \((-2, \frac{8}{5}]\) and \((2, ∞)\).

### Step 6: Solve the Second Inequality
Now, for the second inequality:
\[ \frac{x^2 - 5x + 4}{x^2 - 4} \geq -1 \]
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 gives:
\[ \frac{x(2x - 5)}{(x - 2)(x + 2)} \geq 0 \]

### Step 7: Find Critical Points for Second Inequality
The critical points are:
1. \(x = 0\)
2. \(2x - 5 = 0 \Rightarrow x = \frac{5}{2}\)
3. \(x - 2 = 0 \Rightarrow x = 2\)
4. \(x + 2 = 0 \Rightarrow x = -2\)

### Step 8: Test Intervals for Second Inequality
We test the intervals determined by the critical points:
- For \(x < -2\)
- For \(-2 < x < 0\)
- For \(0 < x < 2\)
- For \(2 < x < \frac{5}{2}\)
- For \(x > \frac{5}{2}\)

After testing these intervals, we find that the inequality holds in the intervals:
- \((-2, 0]\) and \((2, \frac{5}{2}]\) and \((\frac{5}{2}, ∞)\).

### Step 9: Find the Intersection
Now we find the intersection of the two solutions:
1. From the first inequality: \((-2, \frac{8}{5}] \cup (2, ∞)\)
2. From the second inequality: \((-2, 0] \cup (2, \frac{5}{2}] \cup (\frac{5}{2}, ∞)\)

The intersection gives:
\[ (-2, 0] \cup (2, \frac{5}{2}] \]

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