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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​$:(2)\frac {1}{x^2-y^2}=\frac {x(x + y)}{x(x - y)(x + y)^2}$​
​$\frac {1}{x^2+2xy + y^2}=\frac {x(x - y)}{x(x - y)(x + y)^2}$​
​$\frac {1}{x^2+xy}=\frac {(x - y)(x + y)}{x(x - y)(x + y)^2}$​
4
$x^2 + 3x$
-2
a
$2a(2a + 1)$
$6(x + 1)^2$
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​$:(4) \frac {-2(x - 2)}{(x - 2)^2(x + 2)},$​​$\frac {-(x + 2)}{(x - 2)^2(x + 2)}$​