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零五網(wǎng) 全部參考答案 啟東中學(xué)作業(yè)本 2025年啟東中學(xué)作業(yè)本七年級(jí)數(shù)學(xué)上冊(cè)人教版 第9頁(yè)解析答案
1.【閱讀思考】
根據(jù)絕對(duì)值的運(yùn)算性質(zhì)可知,一個(gè)正數(shù)的絕對(duì)值是其本身,一個(gè)負(fù)數(shù)的絕對(duì)值是其相反數(shù),0的絕對(duì)值是0,由此可知,求一個(gè)算式整體的絕對(duì)值,可先判斷整體的正負(fù)性,再求它的絕對(duì)值,最后化簡(jiǎn)。
例如:$|7 + 8| = 7 + 8$,$|5 - 7| = -(5 - 7) = 7 - 5$,$|7 - 4| = 7 - 4$。
【牛刀小試】(1)根據(jù)上述材料,把下列各式去掉絕對(duì)值符號(hào),不要算出最后結(jié)果:
$|3 - 10| = $
10?3
,$|\frac{3}{14} - \frac{3}{17}| = $
$\frac{3}{14}$?$\frac{3}{17}$

【拓展延伸】(2)$|\frac{1}{101} - \frac{1}{100}| + |\frac{1}{102} - \frac{1}{101}| + |\frac{1}{103} - \frac{1}{102}| + … + |\frac{1}{1000} - \frac{1}{999}|$。
解:|$\frac{1}{101}$?$\frac{1}{100}$|+|$\frac{1}{102}$?$\frac{1}{101}$|+|$\frac{1}{103}$?$\frac{1}{102}$|+…+|$\frac{1}{1000}$?$\frac{1}{999}$|
=?($\frac{1}{101}$?$\frac{1}{100}$)?($\frac{1}{102}$?$\frac{1}{101}$)?($\frac{1}{103}$?$\frac{1}{102}$)?…?($\frac{1}{1000}$?$\frac{1}{999}$)
=$\frac{1}{100}$?$\frac{1}{101}$+$\frac{1}{101}$?$\frac{1}{102}$+$\frac{1}{102}$?$\frac{1}{103}$+…+$\frac{1}{999}$?$\frac{1}{1000}$
=$\frac{1}{100}$?$\frac{1}{1000}$
=$\frac{9}{1000}$.
答案:1.(1)10?3 $\frac{3}{14}$?$\frac{3}{17}$
(2)解:|$\frac{1}{101}$?$\frac{1}{100}$|+|$\frac{1}{102}$?$\frac{1}{101}$|+|$\frac{1}{103}$?$\frac{1}{102}$|+…+|$\frac{1}{1000}$?$\frac{1}{999}$|
=?($\frac{1}{101}$?$\frac{1}{100}$)?($\frac{1}{102}$?$\frac{1}{101}$)?($\frac{1}{103}$?$\frac{1}{102}$)?…?($\frac{1}{1000}$?$\frac{1}{999}$)
=$\frac{1}{100}$?$\frac{1}{101}$+$\frac{1}{101}$?$\frac{1}{102}$+$\frac{1}{102}$?$\frac{1}{103}$+…+$\frac{1}{999}$?$\frac{1}{1000}$
=$\frac{1}{100}$?$\frac{1}{1000}$
=$\frac{9}{1000}$.
2.【信息提取】在有些情況下,不需要計(jì)算出結(jié)果也能把絕對(duì)值符號(hào)去掉。
例如:|$\frac{1}{2} - 1$|$ = 1 - \frac{1}{2},$|$\frac{1}{3} - \frac{1}{2}$|$ = \frac{1}{2} - \frac{1}{3},$|$\frac{1}{4} - \frac{1}{3}$|$ = \frac{1}{3} - \frac{1}{4},……【$初步體驗(yàn)】(1)根據(jù)上面的規(guī)律,把下列式子寫(xiě)成去掉絕對(duì)值符號(hào)的形式(不要計(jì)算出結(jié)果):|$\frac{1}{19} - \frac{1}{18}$| =
$\frac{1}{18}-\frac{1}{19}$
;【拓廣應(yīng)用】(2)計(jì)算|$\frac{1}{2} - 1$| + |$\frac{1}{3} - \frac{1}{2}$| + |$\frac{1}{4} - \frac{1}{3}$| + |$\frac{1}{5} - \frac{1}{4}$|;
解:|$\frac{1}{2}$?1|+|$\frac{1}{3}$?$\frac{1}{2}$|+|$\frac{1}{4}$?$\frac{1}{3}$|+|$\frac{1}{5}$?$\frac{1}{4}$|
=1?$\frac{1}{2}$+$\frac{1}{2}$?$\frac{1}{3}$+$\frac{1}{3}$?$\frac{1}{4}$+$\frac{1}{4}$?$\frac{1}{5}$
=1?$\frac{1}{5}$
=$\frac{4}{5}$.
(3)計(jì)算|$\frac{1}{2} - 1$| + |$\frac{1}{3} - \frac{1}{2}$| + |$\frac{1}{4} - \frac{1}{3}$| + … + |$\frac{1}{2023} - \frac{1}{2022}$|。
解:|$\frac{1}{2}$?1|+|$\frac{1}{3}$?$\frac{1}{2}$|+|$\frac{1}{4}$?$\frac{1}{3}$|+…+|$\frac{1}{2023}$?$\frac{1}{2022}$|
=1?$\frac{1}{2}$+$\frac{1}{2}$?$\frac{1}{3}$+$\frac{1}{3}$?$\frac{1}{4}$+…+$\frac{1}{2022}$?$\frac{1}{2023}$
=1?$\frac{1}{2023}$
=$\frac{2022}{2023}$.

答案:2.(1)$\frac{1}{18}$?$\frac{1}{19}$
(2)解:|$\frac{1}{2}$?1|+|$\frac{1}{3}$?$\frac{1}{2}$|+|$\frac{1}{4}$?$\frac{1}{3}$|+|$\frac{1}{5}$?$\frac{1}{4}$|
=1?$\frac{1}{2}$+$\frac{1}{2}$?$\frac{1}{3}$+$\frac{1}{3}$?$\frac{1}{4}$+$\frac{1}{4}$?$\frac{1}{5}$
=1?$\frac{1}{5}$
=$\frac{4}{5}$.
(3)解:|$\frac{1}{2}$?1|+|$\frac{1}{3}$?$\frac{1}{2}$|+|$\frac{1}{4}$?$\frac{1}{3}$|+…+|$\frac{1}{2023}$?$\frac{1}{2022}$|
=1?$\frac{1}{2}$+$\frac{1}{2}$?$\frac{1}{3}$+$\frac{1}{3}$?$\frac{1}{4}$+…+$\frac{1}{2022}$?$\frac{1}{2023}$
=1?$\frac{1}{2023}$
=$\frac{2022}{2023}$.
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