1. Solve the system of equations in the set of real numbers

$$
\left\{\begin{array}{l}
x^{2}+y^{2}+z=2 \\
x+y^{2}+z^{2}=2 \\
x^{2}+y+z^{2}=2
\end{array}\right.
$$
Solution. From the first equation, we subtract the second equation and obtain $x^{2}-x-z^{2}+z=0$, or $(x-z)(x+z-1)=0$. Therefore, $x=z$ or $x=1-z$. Similarly, from the first equation, we subtract the third equation and obtain $y^{2}-y-z^{2}+z=0$, i.e., $(y-z)(y+z-1)=0$. Therefore, $y=z$ or $y=1-z$. There are four possible cases:

(i) $x=z, y=z$. Substituting into the first equation, we get $2 z^{2}+z-1=0$, whose solutions are $z_{1 / 2}=\frac{-1 \pm \sqrt{17}}{4}$. The solutions to the system are

$$
\left(\frac{-1 \pm \sqrt{17}}{4}, \frac{-1 \pm \sqrt{17}}{4}, \frac{-1 \pm \sqrt{17}}{4}\right)
$$

(ii) $x=z, y=1-z$. Substituting into the first equation, we get $2 z^{2}-z-1=0$, whose solutions are $z_{1}=1$ and $z_{2}=-\frac{1}{2}$. The solutions to the system in this case are $(1,0,1)$ and $\left(-\frac{1}{2}, \frac{3}{2},-\frac{1}{2}\right)$.

(iii) $x=1-z, y=z$. Substituting into the first equation of the system, we get $2 z^{2}-z-1=0$, whose solutions are $z_{1}=1$ and $z_{2}=-\frac{1}{2}$. The solutions to the system in this case are $(0,1,1)$ and $\left(\frac{3}{2},-\frac{1}{2},-\frac{1}{2}\right)$.

(iv) $x=1-z, y=1-z$. Substituting into the first equation, we get $2 z^{2}-3 z=0$, whose solutions are $z_{1}=0$ and $z_{2}=\frac{3}{2}$. The solutions to the system in this case are $(1,1,0)$ and $\left(-\frac{1}{2},-\frac{1}{2}, \frac{3}{2}\right)$.