5. Let $k<3, k \neq 0$. Then the conic sections
$$
\frac{x^{2}}{3-k}-\frac{y^{2}}{k}=1 \text { and } \frac{x^{2}}{5}+\frac{y^{2}}{2}=1
$$

must have ( ).
(A) different vertices
(B) different directrices
(C) the same foci
(D) the same eccentricity
5.C.

When $00$, and $3-k>-k$. Then $\frac{x^{2}}{3-k}+\frac{y^{2}}{-k}=1$ represents an ellipse with foci on the $x$-axis. $a^{2}=3-k, b^{2}=-k$. Therefore, $a^{2}-b^{2}=3=c^{2}$, which means the ellipse has the same foci as the given ellipse.