We are asked to solve the inequality $| \frac{x-2}{x+2} | \le 1$.

1. Problem Description

We are asked to solve the inequality $| \frac{x-2}{x+2} | \le 1$.

2. Solution Steps

We can rewrite the inequality as:

$-1 \le \frac{x-2}{x+2} \le 1$

This inequality is equivalent to the following system of inequalities:

$\frac{x-2}{x+2} \le 1$ and $\frac{x-2}{x+2} \ge -1$.

Let's solve the first inequality:

$\frac{x-2}{x+2} \le 1$

$\frac{x-2}{x+2} - 1 \le 0$

$\frac{x-2 - (x+2)}{x+2} \le 0$

$\frac{x-2-x-2}{x+2} \le 0$

$\frac{-4}{x+2} \le 0$

This is equivalent to $x+2 > 0$, since $-4 < 0$.

So, $x > -2$.

Now let's solve the second inequality:

$\frac{x-2}{x+2} \ge -1$

$\frac{x-2}{x+2} + 1 \ge 0$

$\frac{x-2 + (x+2)}{x+2} \ge 0$

$\frac{x-2+x+2}{x+2} \ge 0$

$\frac{2x}{x+2} \ge 0$

We consider two cases:

Case 1: $2x \ge 0$ and $x+2 > 0$.

$x \ge 0$ and $x > -2$. This gives us $x \ge 0$.

Case 2: $2x \le 0$ and $x+2 < 0$.

$x \le 0$ and $x < -2$. This gives us $x < -2$.

Combining the two inequalities we get $x > -2$ and ($x \ge 0$ or $x < -2$).

Since $x > -2$, we must have $x \ge 0$.

3. Final Answer

$x \ge 0$

$[0, \infty)$