1. Given real numbers $a, b, c$ satisfy:
(1) $\sqrt{\frac{a}{b}}(\sqrt{a b}+2 b)=2 \sqrt{a b}+3 b$;
(2) $a=b c$.

Then the number of values for $c$ is ( ).
(A) 1
(B) 2
(C) 3
(D) infinitely many
- 1.C.

Obviously, $ab>0$.
When $a>0, b>0$, from (1) we get $a=3b \Rightarrow c=\frac{a}{b}=3$. When $a<0, b<0$, from (1) we get
$$
a+4 \sqrt{ab}+3b=0,
$$

which simplifies to $\left(\sqrt{\frac{a}{b}}\right)^{2}-4 \sqrt{\frac{a}{b}}+3=0$. Solving this, we get $\sqrt{\frac{a}{b}}=1$ or 3. Therefore, $c=1$ or 9.