1. In the field of real numbers, solve the system of equations

$$
\begin{aligned}
& |1-x|=y+1 \\
& |1+y|=z-2 \\
& |2-z|=x-x^{2}
\end{aligned}
$$
1. The right side of the first equation is a non-negative number, which is reflected in the second equation, where we can remove the absolute value. Also, the right side of the second equation is a non-negative number, which, using the equality $|z-2|=|2-z|$, is reflected in the third equation, where we can remove the absolute value. The given system then has the form

$$
\begin{aligned}
|1-x| & =y+1 \\
1+y & =z-2 \\
z-2 & =x-x^{2}
\end{aligned}
$$

and from this, by simple comparison, we obtain the equation

$$
|1-x|=x-x^{2} .
$$

For $x<1$, we get the equation $1-x=x-x^{2}$, or $(1-x)^{2}=0$, whose solution $x=1$ does not satisfy the assumption $x<1$.

For $x \geq 1$, the equation $x^{2}=1$ results, from which the two solutions $x=-1$ and $x=1$ are obtained, and only $x=1$ satisfies the assumption $x \geq 1$.

From the given system, we then simply calculate the values $y=-1$ and $z=2$. The system thus has the unique solution $(x ; y ; z)=(1 ;-1 ; 2)$.

For a systematic and complete solution, award 6 points. For guessing the solution $(1 ;-1 ; 2)$, award one point.