1. Given positive real numbers $a$ and $b$ satisfy $ab = a + b$. Then $\frac{a}{b} + \frac{b}{a} - ab = (\quad)$.
(A) -2
(B) $-\frac{1}{2}$
(C) $\frac{1}{2}$
(D) 2
$\begin{array}{l}\text { - 1. A. } \\ \frac{a}{b}+\frac{b}{a}-a b=\frac{a^{2}+b^{2}}{a b}-a b \\ =\frac{(a+b)^{2}-2 a b}{a b}-a b \\ =a b-2-a b=-2 .\end{array}$

The translation is as follows:

$\begin{array}{l}\text { - 1. A. } \\ \frac{a}{b}+\frac{b}{a}-a b=\frac{a^{2}+b^{2}}{a b}-a b \\ =\frac{(a+b)^{2}-2 a b}{a b}-a b \\ =a b-2-a b=-2 .\end{array}$