1. If $(2 x-1)(3 x+5)=0$, then $3 x^{2}+\frac{7}{2} x-\frac{3}{2}=(\quad)$.
(A) 1
(B) $-\frac{5}{2}$
(C) $\frac{38}{3}$
(D) 1 or $\frac{38}{3}$
$$
-1 . \mathrm{A} .
$$

The original equation is $6 x^{2}+7 x-5=0$. After dividing by 2 and rearranging, we get
$$
3 x^{2}+\frac{7}{2} x-\frac{3}{2}=1 \text {. }
$$