Find all the values of x that satisfy the following inequation `|(x^2-5x+4)/(x^2-4)|le1`

Options

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

Answer

To solve the inequality \(\left|\frac{x^2 - 5x + 4}{x^2 - 4}\right| \leq 1\), we will break it down step by step.

### Step 1: Rewrite the Inequality
We start with the inequality:
\[ \left|\frac{x^2 - 5x + 4}{x^2 - 4}\right| \leq 1 \]
This can be rewritten as two separate inequalities:
\[ -1 \leq \frac{x^2 - 5x + 4}{x^2 - 4} \leq 1 \]

### Step 2: Solve the Right Inequality
First, we solve the right side:
\[ \frac{x^2 - 5x + 4}{x^2 - 4} \leq 1 \]
This simplifies to:
\[ x^2 - 5x + 4 \leq x^2 - 4 \]
Subtract \(x^2\) from both sides:
\[ -5x + 4 \leq -4 \]
Adding \(4\) to both sides gives:
\[ -5x \leq -8 \]
Dividing by \(-5\) (and reversing the inequality):
\[ x \geq \frac{8}{5} \]

### Step 3: Solve the Left Inequality
Now we solve the left side:
\[ \frac{x^2 - 5x + 4}{x^2 - 4} \geq -1 \]
This simplifies to:
\[ x^2 - 5x + 4 \geq -x^2 + 4 \]
Rearranging gives:
\[ 2x^2 - 5x \geq 0 \]
Factoring out \(x\):
\[ x(2x - 5) \geq 0 \]

### Step 4: Find the Critical Points
The critical points are found by setting each factor to zero:
1. \(x = 0\)
2. \(2x - 5 = 0 \Rightarrow x = \frac{5}{2}\)

### Step 5: Test Intervals
We will test the intervals defined by the critical points \(0\) and \(\frac{5}{2}\):
- For \(x < 0\): Choose \(x = -1\) → \( (-1)(-2) \geq 0 \) (True)
- For \(0 < x < \frac{5}{2}\): Choose \(x = 1\) → \( (1)(-3) \geq 0 \) (False)
- For \(x > \frac{5}{2}\): Choose \(x = 3\) → \( (3)(1) \geq 0 \) (True)

Thus, the solution for \(x(2x - 5) \geq 0\) is:
\[ x \in (-\infty, 0] \cup \left[\frac{5}{2}, \infty\right) \]

### Step 6: Combine the Results
Now we combine the results from both inequalities:
1. From the right inequality: \(x \geq \frac{8}{5}\)
2. From the left inequality: \(x \in (-\infty, 0] \cup \left[\frac{5}{2}, \infty\right)\)

The combined solution is:
\[ x \in \left[\frac{8}{5}, 0\right] \cup \left[\frac{5}{2}, \infty\right) \]

### Final Answer
Thus, the final solution to the inequality \(\left|\frac{x^2 - 5x + 4}{x^2 - 4}\right| \leq 1\) is:
\[ x \in \left[\frac{8}{5}, 2\right) \cup \left[\frac{5}{2}, \infty\right) \]