第93页 - 苏科版八年级数学课课练答案（上下册）

解：原式$=\frac {a-1}{a^2-1}$

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

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

解：原式$=\frac {2(x-2)}{(x+2)(x-2)}-\frac {x-6}{(x+2)(x-2)}$

$ =\frac {2x-4-(x-6)}{(x+2)(x-2)}$

$ =\frac {2x-4-x+6}{(x+2)(x-2)}$

$ =\frac {x+2}{(x+2)(x-2)}$

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

①

解：$\frac {a^2}{a-1}-a-1$

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

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

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

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

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

D

A

$1$