第89页 - 亮点给力提优课时作业本八年级数学江苏版

-2

$\frac {2}{3}$

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

$=\frac {2x}{x+2}$

$=\frac {(x+3)(x-3)}{(x+1)²}÷\frac {x²-x^2+x+3}{x+1}$

$=\frac {(x+3)(x-3)}{(x+1)²} · \frac {x+1}{x+3}$

$= \frac {x-3}{x+1} $

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

又$\frac {a-1}{2}≤1$，所以$a≤3$

又$a $为正整数，$ $且$a-2≠0$，$a-3≠0$，$a+3≠0$，所以$a=1$

则原式$=\frac {1-3}{1+3}=-\frac {1}{2}$

B

$±\frac {1}{2} $

解：$t_{2}＞t_{1}$，理由如下：

由题意，得$t_{1}=\frac {2s}{v}$，$t_{2}= \frac {s}{v+p}+\frac {s}{v-p}=\frac {2sv}{v²-p²}$

∴$\frac {t_{2}}{t_{1}}=\frac {2sv}{v²-p²}÷ \frac {2s}{v}= \frac {2sv}{v²-p²} · \frac {v}{2s} =\frac {v^2}{v²-p^2}$

∵$v＞p＞0$，∴$v²＞v²-p²＞0$

∴$\frac {v^2}{v²-p²}＞1$，即$\frac {t_{2}}{t_{1}}＞ 1$

∴$t_{2}＞t_{1}$