第82页 - 江苏版启东中学作业本八年级数学（上下册）

C

B

x-2

$\frac {1}{x-2}$

$\frac {a}{a-b}$

$=\frac {a²}{b²c} · (-\frac {bc²}{2a}) · \frac {b}{a}$

$=-\frac {c}{2}$

$=\frac {m-3}{2(m-2)}÷\frac {(m+2)(m-2)-5}{m-2}$

$=\frac {m-3}{2(m-2)}·\frac {m-2}{(m+3)(m-3)}$

$=\frac {1}{2m+6}$

$=(\frac {a}{a}-\frac {1}{a})÷(\frac {a^2}{a}-\frac {1}{a})$

$=\frac {a-1}{a}·\frac {a}{a^2-1}$

$=\frac {a-1}{(a-1)(a+1)}$

$=\frac {1}{a+1}$

$=\frac {a+1}{a-1}-\frac {a}{(a-1)^2}·a$

$=\frac {(a+1)(a-1)-a^2}{(a-1)^2}$

$=-\frac {1}{(a-1)^2}$

解：原式$=1-\frac {a-b}{a-2b}·\frac {(a-2b)^2}{(a+b)(a-b)}=1-\frac {a-2b}{a+b}=\frac {a+b}{a+b}-\frac {a-2b}{a+b}=\frac {3b}{a+b}$

当$\frac {a}{b}=\frac {1}{3}$，即$b=3a$时，

原式$=\frac {9a}{a+3a}=\frac {9a}{4a}=\frac {9}{4}$