第4页 - 江苏版启东中学作业本八年级数学（上下册）

$解：(1) \frac {1}{ax}= \frac {b}{abx} ，\frac {1}{bx}= \frac {a}{abx}；(2)\frac {3}{2x^2y}=\frac {9y}{6x^2y^2}，\frac 5{3xy^2}= \frac {10x}{6x^2y^2}；(3) \frac {x-1}{3x^2}=\frac {a(x-1)}{3ax^2}，\frac {2}{ax}=\frac {6x}{3ax^2}$

$解：(4) \frac {a}{2b}=\frac {6a^2}{12ab}，\frac {b}{3a}=\frac {4b^2}{12ab}，\frac {c}{4ab}=\frac {3c}{12ab}；$

$(5)\frac {4a}{5b^2c}= \frac {8a^3c}{10a^2b^2c^2}，\frac {3c}{10a^2b} =\frac {3bc^3}{10a^2b^2c^2}，\frac {5b}{-2ac^2}=-\frac {25ab^3}{10a^2b^2c^2}；$

$(6) \frac {3a}{2a-b}，- \frac {1}{b-2a}=\frac {1}{2a-b}$

$解： \frac {2}{x+1}=\frac {2(x+2)}{(x+1)(x+2)}，\frac 3{x+2}=\frac {3(x+1)}{(x+1)(x+2)}$

$解：\frac {b}{a-x}=\frac {by}{ay-xy}，\frac c{ay-xy}$

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

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

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

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

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

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

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

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

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

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

$\frac {x}{4-2x}= -\frac {x}{2(x-2)} =-\frac {x(x+2)}{2(x+2)(x-2)}$

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

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

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