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第89页

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$\frac {2}{3}$

$解：原式=\frac {a²-4a+4-4}{a-2}×\frac {(a-2)(a+2)}{a-4}$

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

$=a²+2a$

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

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

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

$ 解：原式=\frac {(a-3)²}{a-2}÷\frac {a²-9}{a-2}$

$ =\frac {(a-3)²}{a-2}×\frac {a-2}{(a+3)(a-3)}$

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

$ 因为\frac {a-1}{2}≤1，a为正整数，且a-2≠0，a-3≠0，a+3≠0$

$ 所以a=1$

$ 所以原式=\frac {1-3}{1+3}=-\frac {1}{2}$

$±\frac {1}{2}$

$解：{t}_{1\lt }{t}_2$

$由题意可得$

${t}_1=\dfrac{2s}{v}$

${t}_2=\dfrac{s}{v+p}+\dfrac{s}{v-p}=\dfrac{2vs}{{v}^2-{p}^2}$

$\therefore {t}_1-{t}_2=\dfrac{2s}{v}-\dfrac{2vs}{{v}^2-{p}^2}=\dfrac{-2s{p}^2}{v({v}^2-{p}^2)}$

$\because 0\lt p\lt v$

$\therefore {t}_1-{t}_2\lt 0$

$\therefore {t}_1\lt {t}_2$