1. Given
$$
(\sqrt{3} x+\sqrt{2})^{5}=a_{5} x^{5}+a_{4} x^{4}+a_{3} x^{3}+a_{2} x^{2}+a_{1} x+a_{0} \text {. }
$$

then $\left(a_{5}+a_{3}+a_{1}\right)^{2}-\left(a_{4}+a_{2}+a_{0}\right)^{2}$ is ( ).
(A) 4
(B) 3
(C) 2
(D) 1
1. D.

Let $x=1$, we get
$$
a_{5}+a_{4}+a_{3}+a_{2}+a_{1}+a_{0}=(\sqrt{3}+\sqrt{2})^{5} \text {; }
$$

Let $x=-1$, we get
$$
\begin{array}{l}
-a_{5}+a_{4}-a_{3}+a_{2}-a_{1}+a_{0}=(-\sqrt{3}+\sqrt{2})^{5}, \\
a_{5}-a_{4}+a_{3}-a_{2}+a_{1}-a_{0}=(\sqrt{3}-\sqrt{2})^{5} . \text { (2) }
\end{array}
$$

From equations (1) and (2), we have
$$
\begin{array}{l}
\left(a_{5}+a_{3}+a_{1}\right)^{2}-\left(a_{4}+a_{2}+a_{0}\right)^{2} \\
=\left(a_{5}+a_{4}+a_{3}+a_{2}+a_{1}+a_{0}\right) \cdot \\
\left(a_{5}-a_{4}+a_{3}-a_{2}+a_{1}-a_{0}\right) \\
=(\sqrt{3}+\sqrt{2})^{5}(\sqrt{3}-\sqrt{2})^{5} \\
= {[(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})]^{5}=1 . }
\end{array}
$$