1. Find all triples of real numbers $(x, y, z)$ that satisfy
$$
\left\{\begin{array}{l}
x^{2}-y z=|y-z|+1, \\
y^{2}-z x=|z-x|+1, \\
z^{2}-x y=|x-y|+1 .
\end{array}\right.
$$
1. By symmetry, without loss of generality, assume $x \geqslant y \geqslant z$. Then
$$
\left\{\begin{array}{l}
x^{2}-y z=y-z+1, \\
y^{2}-z x=x-z+1, \\
z^{2}-x y=x-y+1 .
\end{array}\right.
$$
Subtracting (2) from (1) and (3) from (2) respectively, we get
$$
\begin{array}{l}
(x-y)(x+y+z+1)=0, \\
(y-z)(x+y+z-1)=0 .
\end{array}
$$

Thus, $x, y, z$ cannot all be distinct, and they cannot all be the same, otherwise equation (1) would become $0=1$, which is a contradiction. This indicates that exactly two of $x, y, z$ are the same.
We consider two cases:
$$
\begin{array}{l}
\text { (1) } x=y>z, x+y+z=1 ; \\
\text { (2) } x>y=z, x+y+z=-1 \text {. }
\end{array}
$$

By observation, $(x, y, z)$ is a solution if and only if $(-z, -y, -x)$ is a solution. Therefore, without loss of generality, assume
$$
x=y>z, x+y+z=1 \text {. }
$$

In this case, $z=1-2 x$.
Substituting into equation (1), we get
$$
x(3 x-4)=0 \Rightarrow x=0 \text { or } \frac{4}{3} \text {. }
$$

This corresponds to the solutions $(0,0,1)$ and $\left(\frac{4}{3}, \frac{4}{3},-\frac{5}{3}\right)$.
Noting that $x \geqslant y \geqslant z$,
thus, $(x, y, z)$ can only be $\left(\frac{4}{3}, \frac{4}{3},-\frac{5}{3}\right)$.
By cyclic symmetry, there are a total of 6 solutions:
$$
\begin{array}{l}
\left(\frac{4}{3}, \frac{4}{3},-\frac{5}{3}\right),\left(\frac{4}{3},-\frac{5}{3}, \frac{4}{3}\right), \\
\left(-\frac{5}{3}, \frac{4}{3}, \frac{4}{3}\right),\left(-\frac{4}{3},-\frac{4}{3}, \frac{5}{3}\right), \\
\left(-\frac{4}{3}, \frac{5}{3},-\frac{4}{3}\right),\left(\frac{5}{3},-\frac{4}{3},-\frac{4}{3}\right) .
\end{array}
$$