1. Find all real solutions of the system of equations

$$
\frac{1}{x+y}+z=1, \quad \frac{1}{y+z}+x=1, \quad \frac{1}{z+x}+y=1
$$
Solution. If we subtract the first two equations and convert both fractions to a common denominator, we get

$$
\frac{(y+z)-(x+y)}{(x+y)(y+z)}+z-x=0
$$

from which, by factoring out the term $z-x$, we obtain

$$
(z-x)\left(\frac{1}{(x+y)(y+z)}+1\right)=0
$$

By similar manipulations for the differences of other pairs of equations, we also get

$$
(y-z)\left(\frac{1}{(z+x)(x+y)}+1\right)=0, \quad(x-y)\left(\frac{1}{(y+z)(z+x)}+1\right)=0 .
$$

First, assume that the numbers $x, y, z$ are all distinct. Then we can divide the derived equations by their non-zero differences and simplify the system to

$$
\begin{aligned}
& (x+y)(y+z)=-1, \\
& (y+z)(z+x)=-1, \\
& (z+x)(x+y)=-1,
\end{aligned}
$$

which can be rewritten after expansion as

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

The equality $x^{2}=y^{2}=z^{2}$ implies that at least two of the numbers $x, y, z$ are equal, which contradicts the assumption of their distinctness.

In the next case, when $x=y=z$, the original system reduces to solving the equation $1 /(2 x)+x=1$. The quadratic equation $2 x^{2}-2 x+1=0$, which is a consequence of this, has a negative discriminant, and thus the equation $1 /(2 x)+x=1$ has no real solutions.

The remaining case is when exactly two of the numbers $x, y, z$ are the same. Considering the symmetry, it suffices to consider the case $x=y \neq z$. From equation (1), after dividing by the non-zero number $z-x$, we get

$$
-1=(x+y)(y+z)=2 x(x+z) \text {, }
$$

while from the first given equation, we have

$$
\frac{1}{x+y}=\frac{1}{2 x}=1-z
$$

Then we can express the value of $1 /(2 x)$ in two ways as

$$
-x-z=\frac{1}{2 x}=1-z
$$

from which we get $x=-1$ and subsequently $z=3 / 2$. It is easy to verify that the triplet $\left(-1,-1, \frac{3}{2}\right)$ satisfies the original system of equations.

Answer. The system of equations has exactly three solutions $\left(-1,-1, \frac{3}{2}\right),\left(\frac{3}{2},-1,-1\right),\left(-1, \frac{3}{2},-1\right)$.