1. Let $a>b>c>d>0$, and
$$
\begin{array}{l}
x=\sqrt{a b}+\sqrt{c d}, y=\sqrt{a c}+\sqrt{b d}, \\
z=\sqrt{a d}+\sqrt{b c} .
\end{array}
$$

Then the size relationship of $x, y, z$ is ( ).
(A) $x<z<y$
(B) $y<z<x$
(C) $x<y<z$
(D) $z<y<x$
- 1. D.
$$
\begin{array}{l}
\text { From } y^{2}-x^{2}=a c+b d-(a b+c d) \\
=(c-b)(a-d)0,
\end{array}
$$

we know $y<x$.
Similarly, $z<y$.
Therefore, $z<y<x$.