第69页 - 苏科版八年级（初二）数学学习与评价答案（上下册）

4

$3x+x^2$

-2

$a$

$2a(2a+1)$

$6(x+1)²$

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

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

$\frac {2}{6-3x}=-\frac {2}{3(x-2)}$

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

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

$解：\frac {2}{4-x²}=\frac {2(2-x)}{(2+x)(x-2)²}$

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

$解：(1)不合理，\frac {x}{3(y-1)}=\frac {2x}{6(y-1)}，\frac {2}{6-6y}=-\frac {2}{6(y-1)}$