2022年七年级数学上册相反数绝对值的综合应用试题新人教版 .pdf

精编学习资料欢迎下载1 专训 3 数轴、相反数、绝对值的综合应用名师点金：数轴是“数”与“形”结合的工具，有了数轴可以由点读数，也可以由数定点，还可以从几何意义上去理解相反数和绝对值；同时利用数轴可以求相反数，化简绝对值等.总之，这三者之间是相互依存，紧密联系的.点、数对应问题题型 1 数轴上的整数点的问题1.某同学在做数学作业时，不小心将墨水洒在所画的数轴上，如图，被墨水污染部分的整数点有个.（第 1 题）2.在数轴上任取一条长为2 01613个单位长度的线段，则此线段在数轴上最多能盖住的整数点的个数为（）A.2 017 B.2 016 C.2 015 D.2 014 题型 2 数轴上的点表示的数的确定3.已知数轴上点A在原点左边，到原点的距离为8 个单位长度，点 B在原点的右边，从点 A走到点 B，要经过32 个单位长度.（1）求 A，B两点分别表示的数；（2）若点 C 也是数轴上的点，点C到点 B的距离是点C到原点的距离的3 倍，求点C表示的数.精编学习资料欢迎下载2 求值问题题型 1 利用数轴求值4.如图，已知数轴上的点A和点 B分别表示互为相反数的两个数a，b，且 ab，A，B两点间的距离为412，求 a，b 的值.（第 4题）题型 2 绝对值非负性的应用5.已知|15 a|b 12|0，求 2ab 7 的 值.6.当 a 为何值时，|1 a|2 有最小值？并求这个最小值.7.当 a 为何值时，2|4 a|有最大值？并求这个最大值.文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5精编学习资料欢迎下载3 化简问题8.三个有理数a，b，c 在数轴上的对应点的位置如图所示，其中数a，b 互为相反数.试求解以下问题：（第 8 题）（1）判断 a，b，c 的正负性；（2）化简|a b|2a|b|.实际应用问题9.一天上午，出租车司机小王在东西走向的中山路上营运，如果规定向东为正，向西为负，出租车的行车里程如下（单位：千米）：15，3，12，11，13，3，12，18，请问小王将最后一位乘客送到目的地时，一共行驶了多少千米？【导学号：11972022】文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5精编学习资料欢迎下载4 答案112 点拨：被墨水污染部分对应的整数有12，11，10，9，8，10，11，12，13，14，15，16，共 12 个2A3解：(1)A 点表示 的数为 8，B点表示的数为24.(2)由已知得，当点 C在原点左边时，点 C到原点的距离为12 个单位长度；当点 C在原点右边时，点C到原点的距离为6 个单位长度综上所述，点C表示的数为6 或 12.4解：因为a 与 b 互为相反数，所以|a|b|4122 214.又因为 ab，所以 a214，b214.5解：由|15 a|b 12|0，得15a 0，b120，所以a15，b12.所以2a b7215 12 725.6解：当a1 时，|1 a|2 有最小值，这个最小值为2.7解：当a4 时，2|4 a|有最大值，这个最大值为2.8解：(1)a 0，b 0，c0.(2)因为 a，b 互为相反数，所以b a.又因为 a0，b0，所以|a b|2a|b|2a|2a|b|2a2abb.点拨：本题中虽没有标出数轴上原点的位置，但由已知条件a，b 互为相反数，即可确定出原点位置在表示数c 和数 b的两 点之间，从而可以确定出a，b，c 的正负性(2)题化简时，既用到了a，b 的正负性，同时还利用了a，b 互为相反数这一条件9解：|15|3|12|11|13|3|12|18|1531211133121887(千米)答：一共行驶了87 千米点拨：利用绝对值求距离、路程问题中，当出现用“”“”号表示带方向的路程时，求一共行驶的路程时，实际上是求绝对值的和文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 HE10D9K9B3B8 ZD5U6C1N10K5文档编码:CD2R8F10A1V9 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