当前位置:文档之家› 2020年秋人教版七年级上册第一章《有理数》单元测试卷 含答案

2020年秋人教版七年级上册第一章《有理数》单元测试卷 含答案

2020年秋人教版七年级上册第一章《有理数》单元测试卷   含答案
2020年秋人教版七年级上册第一章《有理数》单元测试卷   含答案

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(1)只有符号不同的两个数,我们说其中一个是另一个的相反数;0的相反数还是0; (2)注意: a-b+c 的相反数是-(a-b+c)= -a+b-c ; a-b 的相反数是b-a ; a+b 的相反数是-a-b ; (3)相反数的和为0 ? a+b=0 ? a 、b 互为相反数. (4)相反数的商为-1. (5)相反数的绝对值相等 4.绝对值: (1)正数的绝对值等于它本身,0的绝对值是0,负数的绝对值等于它的相反数; 注意:绝对值的意义是数轴上表示某数的点离开原点的距离; (2) 绝对值可表示为:?????<-=>=) 0a (a )0a (0)0a (a a 或 ???≤-≥=)0()0(a a a a a ; (3) 0a 1a a >?= ; 0a 1a a

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(3) s或15s或30s或45s 【解析】【解答】(2)解:当OI在直线OA的上方时, 有∠MON=∠MOI+∠NOI= (∠AOI+∠BOI))= ∠AOB= ×120°=60°, ∠PON= ×60°=30°, ∵∠MOI=3∠POI, ∴3t=3(30-3t)或3t=3(3t-30), 解得t= 或15; 当OI在直线AO的下方时,

∠MON═(360°-∠AOB)═ ×240°=120°, ∵∠MOI=3∠POI, ∴180°-3t=3(60°- )或180°-3t=3( -60°), 解得t=30或45, 综上所述,满足条件的t的值为 s或15s或30s或45s 【分析】(1)利用角的和差进行计算便可;(2)设,则,,通过角的和差列出方程解答便可;(3)分情况讨论,确定∠MON在不同情况下的定值,再根据角的和差确定t的不同方程进行解答便可. 2.结合数轴与绝对值的知识回答下列问题: (1)探究: ①数轴上表示5和2的两点之间的距离是多少. ②数轴上表示﹣2和﹣6的两点之间的距离是多少. ③数轴上表示﹣4和3的两点之间的距离是多少. (2)归纳: 一般的,数轴上表示数m和数n的两点之间的距离等于|m﹣n|. 应用: ①如果表示数a和3的两点之间的距离是7,则可记为:|a﹣3|=7,求a的值. ②若数轴上表示数a的点位于﹣4与3之间,求|a+4|+|a﹣3|的值. ③当a取何值时,|a+4|+|a﹣1|+|a﹣3|的值最小,最小值是多少?请说明理由. (3)拓展:某一直线沿街有2014户居民(相邻两户居民间隔相同):A1, A2, A3,A4, A5,…A2014,某餐饮公司想为这2014户居民提供早餐,决定在路旁建立一个快餐店P,点P选在什么线段上,才能使这2014户居民到点P的距离总和最小. 【答案】(1)解:①数轴上表示5和2的两点之间的距离是3.

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