1. Given $\left(a^{2}+1\right)\left(b^{2}+1\right)=3(2 a b-1)$. Then the value of $b$ $\left(\frac{1}{a}-a\right)$ is ( ).
(A)0
(B) 1
(C) -2
(D) -1
-1. (D).
From the condition, we have $(a b-2)^{2}+(a-b)^{2}=0$.
Then $a b=2, a=b, a^{2}=2$.
Therefore, $b\left(\frac{1}{a}-a\right)=b \cdot \frac{1-a^{2}}{a}$
$$
=1-a^{2}=1-2=-1 \text {. }
$$