3. Let $a, b$ be positive real numbers and $\frac{1}{a}+\frac{1}{b}-\frac{1}{a-b}=0$. Then, the value of $\frac{b}{a}$ is ( ).
(A) $\frac{1+\sqrt{5}}{2}$
(B) $\frac{1-\sqrt{5}}{2}$ (C) $\frac{-1+\sqrt{5}}{2}$
(D) $\frac{-1-\sqrt{5}}{2}$
3. (C).

From the given, we have $\frac{a+b}{a b}=\frac{1}{a-b}$, which means $a^{2}-b^{2}=a b$. Therefore, $\frac{a}{b}-\frac{b}{a}=1$. Solving this, we get $\frac{b}{a}=\frac{-1+\sqrt{5}}{2}$ (the negative value is discarded).