1. Real numbers $x, y, z$ satisfy $x=y+\sqrt{2}, 2 x y+$ $2 \sqrt{2} z^{2}+1=0$. Then the value of $x+y+z$ is $\qquad$
$=1.0$.
Since $x=y+\sqrt{2}$, we have,
$$
\begin{array}{l}
(x-y)^{2}=2, (x+y)^{2}-4 x y=2, \\
2 x y=\frac{1}{2}(x+y)^{2}-1 .
\end{array}
$$

Substituting this into $2 x y+2 \sqrt{2} z^{2}+1=0$ yields
$$
\frac{1}{2}(x+y)^{2}+2 \sqrt{2} z^{2}=0 \text {. }
$$