第69页

信息发布者：
C
C
$2(x + 1)^{2}$
$(x + 3)(x - 3)^{2}$

$解:\frac {1}{x^2-x}=\frac {1×(x+1)}{(x^2-x)×(x+1)}=\frac{x+1}{x(x-1)(x+1)}$
$\frac {2}{x^2-1}=\frac {2×x}{(x^2-1)×x}=\frac{2x}{x(x-1)(x+1)}$

$解:\frac{1}{a^2b-4b}=\frac{1×(a-2)}{(a^2b-4b)×(a-2)}=\frac{a-2}{b(a-2)^2(a+2)}$
$\frac {1}{b(a-2)^2}=\frac {1×(a+2)}{b(a-2)^2×(a+2)}=\frac{a+2}{b(a-2)^2(a+2)}$

$解:(1)\frac {y}{x(x-y)^2}=\frac {y×y}{x(x-y)^2×y}=\frac {y^2}{xy(x-y)^2}$
$\frac {x}{y(y-x)^2}=\frac {x×x}{y(x-y)^2×x}=\frac {x^2}{xy(x-y)^2}$

$解:\frac {5}{x-9x^2}=\frac {5×(2-3x)^2}{(x-9x^2)×(2-3x)^2}=\frac{5(2x-3)^2}{(2-3x)(2+3x)(2x-3)^2}$
$\frac {2m}{4x^2-12x+9}=\frac {2m×(2-3x)(2+3x)}{(2x-3)^2×(2-3x)(2+3x)}=\frac {2m(2-3x)(2+3x)}{(2-3x)(2+3x)(2x-3)^2}$

$解:\frac {x}{x-y}=\frac {x×(x+y)^2}{(x-y)×(x+y)^2}=\frac {x(x+y)^2}{(x-y)(x+y)^2}$
$\frac {x}{x^2+2xy+y^2}=\frac {x×(x-y)}{(x+y)^2×(x-y)}=\frac {x(x-y)}{(x-y)(x+y)^2}$
$\frac {2}{y^2-x^2}=-\frac {2×(x+y)}{(x^2-y^2)×(x+y)}=\frac {-2(x+y)}{(x-y)(x+y)^2}$

$解:\frac {2}{9-3a}=-\frac {2×(a-3)(a+3)}{3(a-3)×(a-3)(a+3)}=\frac{-2(a-3)(a+3)}{3(a-3)^2(a+3)}$
$\frac {1}{a-6a+9}=\frac {1×3(a+3)}{(a-3)^2×3(a+3)}=\frac{3(a+3)}{3(a-3)^2(a+3)}$
$\frac {2}{3a^2-27}=\frac {2×(a-3)}{3(a^2-9)×(a-3)}=\frac{2(a-3)}{3(a-3)^2(a+3)}$
解：
因为$\frac{A}{x - 1}+\frac{B}{x + 1}=\frac{A(x + 1)+B(x - 1)}{(x + 1)(x - 1)}=\frac{Ax+A + Bx - B}{(x + 1)(x - 1)}=\frac{(A + B)x+A - B}{x^{2}-1}，$
又因为$\frac{3x - 2}{x^{2}-1}=\frac{A}{x - 1}+\frac{B}{x + 1}，$
所以$\frac{3x - 2}{x^{2}-1}=\frac{(A + B)x+A - B}{x^{2}-1}。$
则$\begin{cases}A + B = 3\\A - B=-2\end{cases}，$
将两式相加可得：$2A=1，$解得$A=\frac{1}{2}，$
把$A=\frac{1}{2}$代入$A + B = 3，$得$\frac{1}{2}+B = 3，$解得$B=\frac{5}{2}。$
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