2A. Solve the system of equations in $\mathbb{R}$: $x+\sqrt{y}=3, y+\sqrt{z}=3, z+\sqrt{x}=3$.
Solution. $x-y=\sqrt{z}-\sqrt{y}, y-z=\sqrt{x}-\sqrt{z}, z-x=\sqrt{y}-\sqrt{x}$. Let's assume that $x \leq y$. Then $\sqrt{z}-\sqrt{y} \leq 0$, i.e., $z \leq y$. Further, it follows that $\sqrt{x}-\sqrt{z} \geq 0$, i.e., $x \geq z$, and thus $y \leq x$. Therefore, $x=y$. Similarly, we get $x=z$. Hence, $x=y=z$. Therefore, it is sufficient to solve the equation $x+\sqrt{x}=3$, i.e., $\sqrt{x}=3-x$, where $x \in[0,3]$.

Its solution is $\frac{7-\sqrt{13}}{2}$, so the solution to the system is

$$
\left(\frac{7-\sqrt{13}}{2}, \frac{7-\sqrt{13}}{2}, \frac{7-\sqrt{13}}{2}\right)
$$