1. Let real numbers $a, b, c$ satisfy
$$
\begin{array}{l}
a+b+c=3, a^{2}+b^{2}+c^{2}=4 . \\
\text { Then } \frac{a^{2}+b^{2}}{2-c}+\frac{b^{2}+c^{2}}{2-a}+\frac{c^{2}+a^{2}}{2-b}=(\quad) \text {. }
\end{array}
$$
(A) 0
(B) 3
(C) 6
(D) 9
-、1. D.
From the given, we have
$$
\begin{array}{l}
\frac{a^{2}+b^{2}}{2-c}+\frac{b^{2}+c^{2}}{2-a}+\frac{c^{2}+a^{2}}{2-b} \\
=\frac{4-c^{2}}{2-c}+\frac{4-a^{2}}{2-a}+\frac{4-b^{2}}{2-b}=9 .
\end{array}
$$