5. Given $x, y, z \in \mathbf{R}_{+}$, and $\sqrt{x^{2}+y^{2}}+z=1$. Then the maximum value of $x y+2 x z$ is $\qquad$ .
5. $\frac{\sqrt{3}}{3}$.

From $x^{2}+y^{2}=(1-z)^{2}$, we know
$$
x^{2}=(1-z+y)(1-z-y) \text {. }
$$

Therefore, $x^{2}(y+2 z)^{2}$
$$
\begin{array}{l}
=\frac{1}{3}(1-z+y)(3-3 z-3 y)(y+2 z)(y+2 z) \\
\leqslant \frac{1}{3}\left[\frac{(1-z+y)+(3-3 z-3 y)+(y+2 z)+(y+2 z)}{4}\right]^{4} \\
=\frac{1}{3} .
\end{array}
$$

Thus, $x y+2 x z \leqslant \frac{\sqrt{3}}{3}$.
The equality holds if and only if $x=\frac{\sqrt{3}}{3}, y=z=\frac{1}{3}$.