1. Given real numbers $x, y$ satisfy $x^{3}+y^{3}=2$. Then the maximum value of $x+y$ is $\qquad$ .
Ni.1.2.
Let $x+y=k$, it is easy to know that $k>0$.
From $x^{3}+y^{3}=2$, we get
$$
(x+y)\left(x^{2}-x y+y^{2}\right)=2 \text {. }
$$

Thus, $x y=\frac{1}{3}\left(k^{2}-\frac{2}{k}\right)$.
From this, we know that $x, y$ are the two real roots of the equation about $t$
$$
t^{2}-k t+\frac{1}{3}\left(k^{2}-\frac{2}{k}\right)=0
$$

Therefore, we have
$$
\Delta=k^{2}-\frac{4}{3}\left(k^{2}-\frac{2}{k}\right) \geqslant 0 \text {. }
$$

Solving this, we get $k \leqslant 2$.
Hence, the maximum value of $x+y$ is 2.