第61页 - 苏科版八年级伴你学数学答案（上下册）

C

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

$=1$

$=\frac {b(a+b)-ab}{(a+b)(a-b)}$

$ =\frac {b^2}{a^2-b^2}$

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

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

$解：(1)n ·\frac n{n+1}=n-\frac {n}{n+1}$

$ (2)右边=\frac {n(n+1)-n}{n+1}=\frac {n^2+n-n}{n+1}=\frac {n^2}{n+1}=n ·\frac n{n+1}=左边$

∴等式正确

A

$\frac {5}{3}$

2

-1

$ =\frac {2x-3}{x^3}$

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

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

$=\frac {2(x+3)+(1-x)(x-3)-6×2}{2(x+3)(x-3)}$

$=\frac {2x+6+x-3-x^2+3x-12}{2(x+3)(x-3)}$

$=\frac {-x^2+6x-9}{2(x+3)(x-3)}$

$=-\frac {x-3}{2x+6}$

$=\frac {1-x+1+x}{1-x^2}+\frac 2{1+x^2}+\frac 4{1+x^4}$

$ =\frac {2(1+x^2)+2(1-x^2)}{1-x^4}+\frac 4{1+x^4}$

$ =\frac {4(1+x^4)+4(1-x^4)}{1-x^8}$

$ =\frac 8{1-x^8}$