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8. 根据规划设计,某市工程队准备在开发区修建一条长1120m的盲道,由于采用新的施工方式,实际每天修建的长度比原计划增加10m,从而缩短了工期.假设原计划每天修建盲道$x$m,那么实际修建这条盲道的工期比原计划缩短了
$\frac{11200}{x(x+10)}$
天.答案
$\frac{11200}{x(x+10)}$
解析
原计划工期为$\frac{1120}{x}$天，实际每天修建$(x + 10)$米，实际工期为$\frac{1120}{x + 10}$天，缩短的工期为$\frac{1120}{x} - \frac{1120}{x + 10} = \frac{1120(x + 10) - 1120x}{x(x + 10)} = \frac{11200}{x(x + 10)}$天。
$\frac{11200}{x(x+10)}$
$\frac{11200}{x(x+10)}$
9. 计算:
(1)$\left(m+2+\frac{5}{2-m}\right)\cdot \frac{2m-4}{3-m}$;
(2)$\left(\frac{x+2}{x^{2}-2x}-\frac{x-1}{x^{2}-4x+4}\right)÷ \frac{x-4}{x}$.
(1)$\left(m+2+\frac{5}{2-m}\right)\cdot \frac{2m-4}{3-m}$;
(2)$\left(\frac{x+2}{x^{2}-2x}-\frac{x-1}{x^{2}-4x+4}\right)÷ \frac{x-4}{x}$.
答案
(1)
$\begin{aligned}原式 = \left(m+2+\frac{5}{2-m}\right)\cdot \frac{2m-4}{3-m} \\= \left(\frac{(m+2)(2-m)}{2-m}+\frac{5}{2-m}\right)\cdot \frac{2(m-2)}{3-m} \\= \frac{4-m^2+5}{2-m} \cdot \frac{2(m-2)}{3-m} \\= \frac{9-m^2}{2-m} \cdot \frac{2(m-2)}{3-m} \\= \frac{(3+m)(3-m)}{2-m} \cdot \frac{2(m-2)}{3-m} \\= -2(m+3) \\= -2m-6\end{aligned}$
(2)
$\begin{aligned}原式 = \left(\frac{x+2}{x^{2}-2x}-\frac{x-1}{x^{2}-4x+4}\right)÷ \frac{x-4}{x} \\= \left(\frac{x+2}{x(x-2)}-\frac{x-1}{(x-2)^2}\right) \cdot \frac{x}{x-4} \\= \frac{(x+2)(x-2)-(x-1)x}{x(x-2)^2} \cdot \frac{x}{x-4} \\= \frac{x^2-4-x^2+x}{x(x-2)^2} \cdot \frac{x}{x-4} \\= \frac{x-4}{x(x-2)^2} \cdot \frac{x}{x-4} \\= \frac{1}{(x-2)^2}\end{aligned}$
$\begin{aligned}原式 = \left(m+2+\frac{5}{2-m}\right)\cdot \frac{2m-4}{3-m} \\= \left(\frac{(m+2)(2-m)}{2-m}+\frac{5}{2-m}\right)\cdot \frac{2(m-2)}{3-m} \\= \frac{4-m^2+5}{2-m} \cdot \frac{2(m-2)}{3-m} \\= \frac{9-m^2}{2-m} \cdot \frac{2(m-2)}{3-m} \\= \frac{(3+m)(3-m)}{2-m} \cdot \frac{2(m-2)}{3-m} \\= -2(m+3) \\= -2m-6\end{aligned}$
(2)
$\begin{aligned}原式 = \left(\frac{x+2}{x^{2}-2x}-\frac{x-1}{x^{2}-4x+4}\right)÷ \frac{x-4}{x} \\= \left(\frac{x+2}{x(x-2)}-\frac{x-1}{(x-2)^2}\right) \cdot \frac{x}{x-4} \\= \frac{(x+2)(x-2)-(x-1)x}{x(x-2)^2} \cdot \frac{x}{x-4} \\= \frac{x^2-4-x^2+x}{x(x-2)^2} \cdot \frac{x}{x-4} \\= \frac{x-4}{x(x-2)^2} \cdot \frac{x}{x-4} \\= \frac{1}{(x-2)^2}\end{aligned}$
10.(1)先化简,再求值:$\left(1-\frac{2}{a+1}\right)÷ \frac{a^{2}-2a+1}{a+1}$,其中$a= \frac{4}{3}$;
(2)先化简,再求值:$\left(\frac{x-1}{x}-\frac{x-2}{x+1}\right)÷ \frac{2x^{2}-x}{x^{2}+2x+1}$,其中$x满足x^{2}-2x-2= 0$.
(2)先化简,再求值:$\left(\frac{x-1}{x}-\frac{x-2}{x+1}\right)÷ \frac{2x^{2}-x}{x^{2}+2x+1}$,其中$x满足x^{2}-2x-2= 0$.
答案
(1) $3$；(2) $\frac{1}{2}$
解析
(1) $\left(1-\frac{2}{a+1}\right)÷ \frac{a^{2}-2a+1}{a+1}$
$=\left(\frac{a+1}{a+1}-\frac{2}{a+1}\right)÷\frac{(a-1)^2}{a+1}$
$=\frac{a-1}{a+1}×\frac{a+1}{(a-1)^2}$
$=\frac{1}{a-1}$
当$a=\frac{4}{3}$时，原式$=\frac{1}{\frac{4}{3}-1}=\frac{1}{\frac{1}{3}}=3$
(2) $\left(\frac{x-1}{x}-\frac{x-2}{x+1}\right)÷ \frac{2x^{2}-x}{x^{2}+2x+1}$
$=\left[\frac{(x-1)(x+1)}{x(x+1)}-\frac{x(x-2)}{x(x+1)}\right]÷\frac{x(2x-1)}{(x+1)^2}$
$=\frac{x^2-1-x^2+2x}{x(x+1)}×\frac{(x+1)^2}{x(2x-1)}$
$=\frac{2x-1}{x(x+1)}×\frac{(x+1)^2}{x(2x-1)}$
$=\frac{x+1}{x^2}$
因为$x^2-2x-2=0$，所以$x^2=2x+2$
原式$=\frac{x+1}{2x+2}=\frac{x+1}{2(x+1)}=\frac{1}{2}$
