第77页 - 江苏版启东中学作业本八年级数学（上下册）

3

5或10

2(x+1)或2(x²-1)

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

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

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

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

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

$\frac {1}{2a+4}=\frac {1}{2(a+2)}=\frac {(a-2)^2}{2(a+2)(a-2)^2}$

解：$\frac {1}{8x-4y}=-\frac {y+2x}{4(y+2x)(y-2x)}$

$\frac {1}{4y-8x}=\frac {y+2x}{4(y+2x)(y-2x)}$

$\frac {3x}{y^2-4x^2} =\frac {12x}{4(y+2x)(y-2x)}$

解：两个分式分母的公因式$a=x-1$，

最简公分母$b=3(x+1)(x-1)$

∴$ \frac {b}{a}=\frac {3(x+1)(x-1)}{x-1}=3(x+1)=3$，解得$x=0$

∴$\frac {1}{3x^2-3}=-\frac {1}{3}$

$\frac {2}{x-1}=-2$

解：例如，取$a=1$，$b=2$，$c=3$，$d=6$

有$\frac {1}{2}=\frac {3}{6}$

则$(1)\frac {1}{3}=\frac {2}{6}$；$(2)\frac {1+2}{2}=\frac {3+6}{6}=\frac {3}{2}$；$(3)\frac {1+2}{1-2}=\frac {3+6}{3-6}=-3$

观察发现各组中的两个分式相等，选择第$(2)$组进行说明

∵$a$、$b$、$c$、$d$都不等于$0$，且$\frac {a}{b}=\frac {c}{d}$

∴$\frac {a}{b}+1=\frac {c}{d}+1$，即$\frac {a+b}{b}=\frac {c+d}{d}$