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$ \begin{aligned}解:原式&=- \frac{a}{bc^2}•\frac{bc}{2a} \\ &=- \frac{1}{2c} \\ \end{aligned}$
$ \begin{aligned}解:原式&=\frac{2a-(a+1)}{a-1} \\ &=\frac{a-1}{a-1} \\ &=1 \\ \end{aligned}$
$解:原式 = \frac{m-15}{(m+3)(m-3)}+ \frac{2(m+3)}{(m+3)(m-3)}\ $
$\hspace{1.4cm}= \frac{m-15+2m+6}{(m+3)(m-3)}$
$\hspace{1.4cm}=\frac{3(m-3)}{(m+3)(m-3)}$
$\hspace{1.4cm}=\frac{3}{m+3}$
$ \begin{aligned}解:原式&=\frac{a-2}{a-1}•\frac{a(a-1)}{(a-2)^2} \\ &=\frac {a}{a-2} \\ \end{aligned}$
$ 解: 原式= \frac{a^2+2a+1-5-2a}{a+1}•\frac {a+1}{(a+2)^2} $
$\hspace{1.4cm}= \frac{(a+2)(a-2)}{(a+1)}•\frac{a+1}{(a+2)^2}$
$\hspace{1.4cm}=\frac{a-2}{a+2}. $
$∵a=\sqrt 9+|-2|-(\frac{1}{2})^{-1}=3+2-2=3，∴原式=\frac{3-2}{3+2}=\frac{1}{5}.$
$解:(1)观察规律可得\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}.$
$ (2)∵\frac{1}{n+1}+\frac{1}{n(n+1)}=\frac{n}{n(n+1)}+\frac{1}{n(n+1)}=\frac{n+1}{n(n+1)}=\frac 1n，$
$∴\frac{1}{n}=\frac{1}{n+1}+\frac{1}{n(n+1)}.$
$解:(1)由已知可得\begin{cases}{a(v_甲+v_乙)=a，}\\{b(v_甲-v_乙)=a，}\end{cases}$
$∴v_甲=\frac {a+b}{2b}，v_乙=\frac {b-a}{2b}，$
$(2)由(1)得\frac {v_甲}{v_乙}=\frac{a+b}{b-a}=\frac{7}{3}，$
$∴\frac{a}{b}=\frac{2}{5}.$
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