第24页 - 启东中学作业本八年级数学人教版

$解:原式=-\frac {3a}{4b}×\frac {1}{9a^{2}b}$

$=-\frac{1}{12ab^{2}}$

$解:原式=\frac {2(x-3)}{(x-2)^{2}}×\frac {(x-2)(x+3)}{3-x}$

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

$ 解:原式=\frac {(x-3)^{2}}{x}×\frac {x(x+3)}{(x+3)(x-3)}$

$=x-3$

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

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

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

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

$ 解:原式=\frac {(3x-y)^{2}}{(x+2y)(x-2y)}×\frac {x+2y}{3x-y}$

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

$解:原式=\frac {3}{a(a+1)}×a(a-1)×\frac {a}{a-1}$

$=\frac{3a}{a+1}$

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

$=a-b$

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

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

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

$=-y$