含绝对值的不等式解法练习题及答案.pdf

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1、学习必备欢迎下载例 1 不等式|83x|0 的解集是 A BRCx|x D8383分析，即|83x|083x0 x83答选 C例 2 绝对值大于2 且不大于5 的最小整数是 A 3 B 2 C 2 D 5 分析列出不等式解根据题意得2|x|5从而 5x 2 或 2x5，其中最小整数为5，答选 D例 3不等式 4|13x|7 的解集为 _分析利用所学知识对不等式实施同解变形解原不等式可化为4|3x1|7，即 43x17 或 7 解之得 或 ，即所求不等式解集为 或 3x14x2x1x|2x1x53835383例 4已知集合Ax|2|62x|5，xN，求 A分析转化为解绝对值不等式解2|62x|5

2、 可化为2|2x6|5 即 ，或，52x652x622x62即，或，12x112x82x4解之得 或 4xx211212因为 xN，所以 A0，1，5 说明：注意元素的限制条件例 5 实数 a，b 满足 ab0，那么 -第 1 页，共 5 页精品p d f 资料 可编辑资料-学习必备欢迎下载A|ab|a|b|B|ab|ab|C|ab|ab|D|ab|a|b|分析根据符号法则及绝对值的意义解a、b 异号，|ab|ab|答选 C例 6 设不等式|xa|b 的解集为 x|1 x2，则 a，b 的值为 A a1，b 3 Ba 1，b3 Ca 1，b 3 Dab，1232分析解不等式后比较区间的端点解由

3、题意知，b0，原不等式的解集为x|a bx ab，由于解集又为x|1x 2 所以比较可得ab1ab2ab ，解之得，1232答选 D说明：本题实际上是利用端点的位置关系构造新不等式组例 7 解关于 x 的不等式|2x1|2m 1(m R)分析分类讨论解 若 即，则 恒不成立，此时原不等2m10m|2x1|2m112式的解集为；若 即，则 ，所以2m10m(2m1)2x12m11m12xm综上所述得：当时原不等式解集为；当时，原不等式的解集为mm1212x|1 mxm 说明：分类讨论时要预先确定分类的标准例解不等式8 3212|xx分析一般地说，可以移项后变形求解，但注意到分母是正数，所以能直接

4、去分母-第 2 页，共 5 页精品p d f 资料 可编辑资料-文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9

5、 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6

6、S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H1

7、0 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R

8、3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7

9、I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码

10、:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1学习必备欢迎下载解注意到分母|x|20，所以原不等式转化为2(3|x|)|x|2，整理得|x|xx|x，

11、从而可以解得，解集为 4343434343说明：分式不等式常常可以先判定一下分子或者分母的符号，使过程简便例 9 解不等式|6|2x1|1分析以通过变形化简，把该不等式化归为|ax b|c 或|axb|c 型的不等式来解解事实上原不等式可化为6|2x1|1 或6|2x1|1 由得|2x1|5，解之得 3x2；由得|2x1|7，解之得x3 或 x 4从而得到原不等式的解集为x|x 4 或 3x2 或 x3 说明：本题需要多次使用绝对值不等式的解题理论例 10已知关于x 的不等式|x2|x3|a 的解集是非空集合，则实数 a的取值范围是 _分析可以根据对|x2|x3|的意义的不同理解，获得多种方法

12、解法一当 x 2 时，不等式化为x2 x3a 即 2x1a 有解，而2x15，a5当 2x3 时，不等式化为x2 x3a 即 a5当 x3 是，不等式化为x2x3a 即 2x1 a有解，而 2x 15，a5综上所述：a 5 时不等式有解，从而解集非空解法二|x2|x3|表示数轴上的点到表示2 和 3 的两点的距离之和，显然最小值为3(2)5故可求 a 的取值范围为a5解法三利用|m|n|mn|得|x2|x3|(x2)(x3)|5所以 a 5时不等式有解说明：通过多种解法锻炼思维的发散性例 11 解不等式|x1|2x分析一对 2x 的取值分类讨论解之解法一原不等式等价于：或 2x0 x12xx1

13、x2或 2x0 xR-第 3 页，共 5 页精品p d f 资料 可编辑资料-文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z

14、1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R

15、9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D

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17、10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6

18、R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I

19、7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1学习必备欢迎下载由得或 x2x1212即，所以；x2xx21212由得 x2综合得所以不等

20、式的解集为xx|x1212分析二利用绝对值的定义对|x1|进行分类讨论解之解法二因为|x1|x1x1x1x1，原不等式等价于：或xxxxxx10121012由得即；xx11212x由得 即 x112x所以不等式的解集为x|x12例 12 解不等式|x 5|2x3|1分析设法去掉绝对值是主要解题策略，可以根据绝对值的意义分区间讨论，事实上，由于 时，时 x5|x5|0 x|2x3|032所以我们可以通过，将 轴分成三段分别讨论325x解 当 时，所以不等式转化为xx502x3032(x5)(2x3)1，得 x 7，所以 x 7；-第 4 页，共 5 页精品p d f 资料 可编辑资料-文档编码:

21、CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C

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24、9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10

25、 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3

26、A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I

27、1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1学习必备欢迎下载当 时，同理不等式化为32x5(x5)(2x3)1，解之得，所以 ；xx51313当 x5 时，原不等式可化为x5(2x3)1，解之得 x 9，所以 x5综上所述

28、得原不等式的解集为或 x|xx713说明：在含有绝对值的不等式中，“去绝对值”是基本策略例 13 解不等式|2x1|2x 3|分析本题也可采取前一题的方法：采取用零点分区间讨论去掉绝对值，但这样比较复杂如果采取两边平方，即根据解|a|b|ab22之，则更显得流畅，简捷解原不等式同解于(2x 1)2(2x3)2，即 4x24x14x212x9，即 8x8，得 x1所以原不等式的解集为x|x 1 说明：本题中，如果把2x 当作数轴上的动坐标，则|2x1|2x3|表示2x到 1 的距离大于2x 到 3 的距离，则2x 应当在 2 的右边，从而2x2 即 x1-第 5 页，共 5 页精品p d f 资

29、料 可编辑资料-文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7

30、I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码

31、:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1

32、C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9

33、 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6

34、S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1文档编码:CW8Z1C5H2R9 HD6D6S9P2H10 ZG6R3A10I7I1