(2) Let $a, b$ be positive real numbers, $A=\frac{a+b}{2}, B=\frac{2}{\frac{1}{a}+\frac{1}{b}}$. If $A+B=a-b$, then the value of $\frac{a}{b}$ is $\qquad$
(2) $3+2 \sqrt{3}$ Hint: From the given, we have $\frac{a+b}{2}+\frac{2}{\frac{1}{a}+\frac{1}{b}}=a-b$. Simplifying, we get $a^{2}-$ $6 a b-3 b^{2}=0$, which is $\left(\frac{a}{b}\right)^{2}-6 \cdot \frac{a}{b}-3=0$. Solving, we get $\frac{a}{b}=3+2 \sqrt{3}$ (the negative root has been discarded).