1. Let $a^{2}+1=3 a, b^{2}+1=3 b$, and $a \neq b$, then the value of the algebraic expression $\frac{1}{a^{2}}+\frac{1}{b^{2}}$ is ( ).
(A) 5
(B) 7
(C) 9
(D) 11
-、1.B.
From the given conditions, we have
$$
a^{2}-3 a+1=0, b^{2}-3 b+1=0,
$$

and $a \neq b$, so $a$ and $b$ are the two roots of the quadratic equation $x^{2}-3 x+$ $1=0$. Therefore,
$$
\begin{array}{l}
a+b=3, a b=1 . \\
\text { Hence } \frac{1}{a^{2}}+\frac{1}{b^{2}}=\frac{a^{2}+b^{2}}{a^{2} b^{2}}=\frac{(a+b)^{2}-2 a b}{(a b)^{2}}=7 .
\end{array}
$$