1. Find all real numbers $x$ for which the inequality

$$
|||2-x|-x|-8| \leq 2008
$$

holds.
I/1. First, we distinguish two cases based on the value of the expression $2-x$. If $x \geq 2$, then $|2-x|=-(2-x)$, and the inequality can be rewritten as $|x-2-x|-8| \leq 2008$, which is equivalent to $6 \leq 2008$. Therefore, all numbers $x$ that are greater than or equal to 2 satisfy the inequality.

Let's consider $x<2$ and $x \leq 2$ or $x \geq 2$ and $x \leq 2$ ) .. $\mathbf{1}$ point

![](https://cdn.mathpix.com/cropped/2024_06_07_24964e3bc413fb5af60fg-06.jpg?height=57&width=1633&top_left_y=1722&top_left_x=220)
Conclusion that the inequality holds for $1 \leq x<2$ (or $1<x<2$ or $1 \leq x \leq 2$ ) ............ 2 points Conclusion that the inequality is satisfied for $-1007 \leq x \leq 1$ (or $-1007 \leq x<1$ )....... 2 points Conclusion that all numbers $x \geq-1007$ satisfy the inequality 1 point