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第91页

信息发布者：

$ C$

$2(x+1)^2$

$-\frac{9}{2}$

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

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

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

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

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

$\frac {x}{y(y-x)^2}=\frac {x^2}{xy(x-y)^2}$

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

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

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

$\frac {x}{x^2+2xy+y^2}=\frac {x(x-y)}{(x-y)(x+y)^2}，$

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

解：$\frac {2}{9-3a}=\frac {-2(a-3)(a+3)}{3(a-3)^2(a+3)}，$

$\frac {1}{a^2-6a+9}=\frac {3(a+3)}{3(a-3)^2(a+3)}，$

$\frac {2}{3a^2-27}=\frac {2(a-3)}{3(a-3)^2(a+3)}$

解：$∵\frac{A}{x-1}+\frac{B}{x+1}=\frac{A(x+1)+B(x-1)}{(x+1)(x-1)}=\frac{Ax+A+Bx-B}{(x+1)(x-1)}=\frac{(A+B)x+A-B}{x^2-1}$

又$∵\frac{3x-2}{x^2-1}=\frac{A}{x-1}+\frac{B}{x+1}$

$∴\frac{3x-2}{x^2-1}=\frac{(A+B)x+A-B}{x^2-1}$

$∴\begin{cases} A+B=3 \\ A-B=-2 \end{cases}$

解得$\begin{cases} A=\frac{1}{2} \\ B=\frac{5}{2} \end{cases}$