1. If $x=\frac{a-b}{a+b}$, and $a \neq 0$, then $\frac{b}{a}$ equals ( ).
(A) $\frac{1-x}{1+x}$
(B) $\frac{1+x}{1-x}$
(C) $\frac{x-1}{x+1}$
(D) $\frac{x+1}{x-1}$
- 1.A.

Since $a \neq 0$, we have $x=\frac{a-b}{a+b}=\frac{1-\frac{b}{a}}{1+\frac{b}{a}}$. Let $\frac{b}{a}=y$, then $x=\frac{1-y}{1+y}$.
Also, since $a \neq 0$, then $x \neq -1$. Therefore, $y=\frac{1-x}{1+x}$.