1. If $a^{2}+b^{2}=4$, then
$$
\begin{array}{l}
\sqrt[3]{a(b-4)}+\sqrt{a b-3 a+2 b-6} \\
=(\quad)
\end{array}
$$
(A) -2
(B) 0
(C) 2
(D) 4
- 1. C.

From the condition $a^{2}+b^{2}=4$, we know $-2 \leqslant a, b \leqslant 2$. Therefore, $a+2 \geqslant 0$, and $b-3<0$.
Thus, $a b-3 a+2 b-6$
$$
=(a+2)(b-3) \leqslant 0 \text{. }
$$

Also, from the meaningfulness of the square root, we have
$$
a b-3 a+2 b-6 \geqslant 0 \text{. }
$$

Therefore, $a=-2, b=0$.
Hence $\sqrt[3]{a(b-4)}+\sqrt{a b-3 a+2 b-6}=2$.