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第83页

信息发布者：

解:把几个异分母的分式分别化成与原来的分式相等的同分母的分式，叫做分式的通分。

解$：(1)\frac {1}{3a^2b}=\frac {4b}{12a^2b^2}，$

$\frac {1}{4ab^2}=\frac {3a}{12a^2b^2}，$

$\frac {1}{12ab}=\frac {ab}{12a^2b^2}$

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

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

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

$x^2 + 3x$

-2

$2a(2a + 1)$

$6(x + 1)^2$

解$：(1) \frac {2b}{2abc}，$$\frac {c}{2abc}$

解$：(2) \frac {9}{3(x - 2)}，$$\frac {-2}{3(x - 2)}$

解$：(3) \frac {x - 4}{(x + 1)(x - 4)^2}，$$\frac {2(x + 1)}{(x + 1)(x - 4)^2}$

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