第65页

信息发布者：
C
D
$4a^2y$
$6ax^2$
$3xy^2$
$(x+3)(x-3)^2$
$解:\frac {1}{x^2-x}=\frac {1×(x+1)}{(x^2-x)×(x+1)}=\frac{x+1}{x(x-1)(x+1)}$
$\frac {2}{x^2-1}=\frac {2×x}{(x^2-1)×x}=\frac{2x}{x(x-1)(x+1)}$
$解:\frac{1}{a^2b-4b}=\frac{1×(a-2)}{(a^2b-4b)×(a-2)}=\frac{a-2}{b(a-2)^2(a+2)}$
$\frac {1}{b(a-2)^2}=\frac {1×(a+2)}{b(a-2)^2×(a+2)}=\frac{a+2}{b(a-2)^2(a+2)}$
$解:(1)\frac {y}{x(x-y)^2}=\frac {y×y}{x(x-y)^2×y}=\frac {y^2}{xy(x-y)^2}$
$\frac {x}{y(y-x)^2}=\frac {x×x}{y(x-y)^2×x}=\frac {x^2}{xy(x-y)^2}$
$解:\frac {5}{x-9x^2}=\frac {5×(2-3x)^2}{(x-9x^2)×(2-3x)^2}=\frac{5(2x-3)^2}{(2-3x)(2+3x)(2x-3)^2}$
$\frac {2m}{4x^2-12x+9}=\frac {2m×(2-3x)(2+3x)}{(2x-3)^2×(2-3x)(2+3x)}=\frac {2m(2-3x)(2+3x)}{(2-3x)(2+3x)(2x-3)^2}$
$解:\frac {x}{x-y}=\frac {x×(x+y)^2}{(x-y)×(x+y)^2}=\frac {x(x+y)^2}{(x-y)(x+y)^2}$
$\frac {x}{x^2+2xy+y^2}=\frac {x×(x-y)}{(x+y)^2×(x-y)}=\frac {x(x-y)}{(x-y)(x+y)^2}$
$\frac {2}{y^2-x^2}=-\frac {2×(x+y)}{(x^2-y^2)×(x+y)}=\frac {-2(x+y)}{(x-y)(x+y)^2}$
$解:\frac{a+1}{a}=\frac{(a+1)^2}{a(a+1)}=\frac{a^2+2a+1}{a(a+1)}~~①，$
$\ \frac{a+2}{a+1}=\frac{a(a+2)}{a(a+1)}=\frac{a^2+2a}{a(a+1)}~~②.\ $
$∵a＞0，∴a(a+1)＞0，$
$通分后发现两个分式的分母相同，①中的分子比②中的分子大1，\ $
$∴\frac{a+1}{a}＞\frac{a+2}{a+1}.$
$解:\frac {2}{9-3a}=-\frac {2×(a-3)(a+3)}{3(a-3)×(a-3)(a+3)}=\frac{-2(a-3)(a+3)}{3(a-3)^2(a+3)}$
$\frac {1}{a-6a+9}=\frac {1×3(a+3)}{(a-3)^2×3(a+3)}=\frac{3(a+3)}{3(a-3)^2(a+3)}$
$\frac {2}{3a^2-27}=\frac {2×(a-3)}{3(a^2-9)×(a-3)}=\frac{2(a-3)}{3(a-3)^2(a+3)}$
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