5. Given that $x, y, z$ are real numbers. If $x^{2}+y^{2}=1$, $y^{2}+z^{2}=2, x^{2}+z^{2}=2$, then the minimum value of $x y+y z+x z$ is ( ).
(A) $\frac{5}{2}$
(B) $\frac{1}{2}+\sqrt{3}$
(C) $-\frac{1}{2}$
(D) $\frac{1}{2}-\sqrt{3}$
5.D.

Given $x^{2}+y^{2}=1, y^{2}+z^{2}=2, x^{2}+z^{2}=2$, we solve to get $x= \pm \frac{\sqrt{2}}{2}, y= \pm \frac{\sqrt{2}}{2}, z= \pm \frac{\sqrt{6}}{2}$.
When $x$ and $y$ have the same sign, and $x$ and $z$ have opposite signs, $x y+y z+x z$ has a minimum value, and the minimum value is
$$
\frac{1}{2}-\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}=\frac{1}{2}-\sqrt{3} \text {. }
$$