第83页 - 亮点给力提优课时作业本八年级数学江苏版

$4a²y$，$6ax²$，$3xy²$

2x+3

6

0或4

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

$(2)$由$(1)$，得$\frac {a-1}{a+1}=\frac {(a-1)(b+1)}{(a+1)(b+1)}$，∴$\frac {a-1}{a+1}=\frac {ab+a-b-1}{ab+a+b+1}$

∵$ab=3$，$a+b=4$

∴$(a-b)²=(a+b)²-4ab=4²-4×3=4$，即$a-b= ±2$

当$a-b = 2 $时，$\frac {a-1}{a+1}=\frac {ab+a-b-1}{ab+a+b+1}=\frac {3+2-1}{3+4+1}=\frac {1}{2}$；

当$a-b=-2$时，$\frac {a-1}{a+1}=\frac {ab+a-b-1}{ab+a+b+1}=\frac {3-2-1}{3+4+1}=0$

综上，$\frac {a-1}{a+1}$的值为$\frac {1}{2}$或$0$

A

$3(2m-3)$或$3(2m+3)$

或$3(2m-3)(2m+3) $

解：$(1)$∵$\frac {a}{b}=\frac {c}{d}$，∴$\frac {a}{b}+1=\frac {c}{d}+1$，即$\frac {a+b}b=\frac {c+d}d$

$(2)$∵$a$，$b$，$c$，$d$都不为$0$，∴可设$\frac {a}{b}= \frac {c}{d}=k(k≠0)$，即$a=bk$，$c=dk$

∴$\frac {a+b}{a-b}=\frac {bk+b}{bk-b}=\frac {b(k+1)}{b(k-1)}=\frac {k+1}{k-1} $，$\frac {c+d}{c-d}=\frac {dk+d}{dk-d}=\frac {d(k+1)}{d(k-1)}=\frac {k+1}{k-1}$

∴$\frac {a+b}{a-b}=\frac {c+d}{c-d}$