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1、第 1 讲实数【考纲要求】1理解有理数、无理数和实数的概念，会用数轴上的点表示有理数2借助数轴理解相反数和绝对值的意义，会求一个数的相反数、倒数与绝对值3理解平方根、算术平方根、立方根的概念，会求一个数的算术平方根、平方根、立方根4理解科学记数法、近似数与有效数字的概念，能按要求用四舍五入法求一个数的近似值，能正确识别一个数的有效数字的个数，会用科学记数法表示一个数5熟练掌握实数的运算，会用各种方法比较两个实数的大小.【命题趋势】实数是中学数学重要的基础知识，中考中多以选择题、填空题和简单的计算题的形式出现，主要考查基本概念、基本技能以及基本的数学思想方法另外，命题者也会利用分析归纳、总结规律
2、等题型考查考生发现问题、解决问题的能力.【考点探究】考点一、实数的分类【例 1】四个数 5，0.1，12，3中为无理数的是()A 5 B 0.1 C12D3 解析：因为 5 是整数属于有理数，0.1 是有限小数属于有理数，12是分数属于有理数，3开不尽方是无理数，故选D 答案：D 方法总结一个数是不是无理数，应先计算或者化简再判断有理数都可以化成分数的形式常见的无理数有四种形式：(1)含有 的式子；(2)根号内含开方开不尽的式子；(3)无限且不循环的小数；(4)某些三角函数式触类旁通1 在实数 5，37，2，4中，无理数是()A5 B 37C2 D4 考点二、相反数、倒数、绝对值与数轴【例 2
3、】(1)15的倒数是 _；(2)(3)2的相反数是()A6 B 6 C9 D 9(3)实数 a，b 在数轴上的位置如图所示，化简|ab|b a2_.解析：(1)15的倒 数为1155；(2)因为(3)29,9 的相反数是 9，故选 D；(3)本题考查了绝对值，平方根及数轴的有关知识由图可知，a 0，b0，|a|b|，所以 a b0，ba0，原式 abba2a.答案：(1)5(2)D(3)2a方法总结1求一个数的相反数，直接在这个数的前面加上负号，有时需要化简得出2解有关绝对值和数轴的问题时常用到字母表示数的思想、分类讨论思想和数形结合思想3相反数是它本身的数只有0；绝对值是它本身的数是0 和正
4、数(即非负数)；倒数是它本身的数是 1.触类旁通2 下列各数中，相反数等于5 的数是()A 5 B5 C15D15考点三、平方根、算术平方根与立方根【例 3】(1)(2)2的算术平方根是()A2 B 2 C 2 D2(2)实数 27 的立方根是 _ 解析：(1)(2)2的算术平方根，即22|2|2；(2)27 的立方根是3273.答案：(1)A(2)3 方法总结1对于算术平方根，要注意：(1)一个正数只有一个算术平方根，它是一个正数；(2)0 的算术平方根是0；(3)负数没有算术平方根；(4)算术平方根a具有双重非负性：被开方数 a 是非负数，即a0；算术平方根a本身是非负数，即a0.2(3a
5、)3a，3a3a.触类旁通3 4 的平方根是()A2 B 2 C16 D 16 考点四、科学记数法、近似数、有效数字【例 4】年安徽省有6 82 000 名初中毕业生参加中考，按四舍五入保留两位有效数字，682 000 用科学记数法表示为()A0.69106B6.82105C0.68106D6.8105解析：用科学记数法表示的数必须满足a10n(1|a|10，n 为整数)的形式；求近似数时注意看清题目要求和单位的换算；查有效数字时，要从左边第1 个非零数查起，到精确到的数为止.682 0006.821056.8105答案：D 方法总结1用 科学记数法表示数，当原数的绝对值大于或等于1 时，n
6、等于原数的整数位数减 1；当原数的绝对值小于1 时，n 是负整数，它的绝对值等于原数中左起第一位非零数字前零的个数2取一个数精确到某一位的近似数时，应对“某一位”后的第一个数进行四舍五入，而之后的数不予考虑3用科学记数法表示的近似数，乘号前面的数(即 a)的有效数字即为该近似数的有效数字；而这个近似数精确到哪一位，应将用科学记数法表示的数还原成原来的数，再看最后一个有效数字处于哪一个数位上触类旁通4 某种细胞的直径是51 04毫米，这个数是()A0.05 毫米B0.005 毫米C0.000 5 毫米D0.000 05 毫米考点五、非负数性质的应用【例 5】若实数 x，y 满足x2(3y)20，
7、则代数式xyx2的值为 _解析：因为x20，(3y)20，而x2(3y)20，所以 x20,3y0，解得 x2，y3，则 xyx223222.答案：2 文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10
8、G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL1
9、0G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL
10、10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:C
11、L10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:
12、CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码
13、:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编
14、码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3方法 总结常见的非负数的形式有三种：|a|，a(a0)，a2，若它们的和为零，则每一个式子都为0.触类旁通5 若|m3|(n2)20，则 m 2n 的值为()A 4 B 1 C0 D 4 考点六、实数的运算【例 6】计算：(1)213cos 30|5|(2 011)0.(2)(1)2 011123 cos 68 50|338sin 60|.(1)分析：2112，cos 3032，|5|5，(2 011)01.解：原式12332 511232516.(2)分析：123(21)3238，cos 68 501，sin 6
15、032解：原式 18133 83283.点拨：(1)根据负整数指数幂的意义可把负整数指数幂转化为正整数指数幂运算，即ap1ap(a0)(2)a0 1(a0)方法总结提高实数的运算能力，首先要认真审题，理解有关概念；其次要正确、灵活地应用零指数、负整数指数的定义、特殊角的三角函数、绝对值、相反数、倒数等相关知识及实数的六种运算法则，根据运算律及顺序，选择合理、简捷的解题途径要特别注意把好符号关考点七、实数的大小比较【例 7】比较 2.5，3，7的大小，正确的是()A 32.57 B2.5 37 C 372.5 D72.5 3 解析：由负数小于正数可得3最小，故只要比较2.5 和7的大小即可，由2
16、.52(7)2，得2.57，所以 32.57.答案：A 方法总结实数的各种比较方法，要明确应用条件及适用范围如：“差值比较法”用于比较任意两数的大小，而“商值比较法”一般适用于比较符号相同的两个数的大小，还有“平方法”、“倒数法”等要依据数值特点确定合适的方法触类旁通6 在 6,0,3,8 这四个数中，最小的数是()A 6 B0 C3 D8【经典考题】1(2013 黄石)13的倒数是()A13B3 C 3 D132(2013 南京)下列四个数中，负数是()A|2|B(2)2C2 D223(2013 北 京)首届中国(北京)国际服 务贸易交易会(京交会)于年 6 月 1 日闭幕，本届京交会期间签
17、订的项目成交总金额达60 110 000 000 美元将 60 110 000 000 用科学记数法表示应为()A6.011109B60.11109C6.0111010D0.60111011文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 Z
18、C6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10
19、ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10
20、 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z1
21、0 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z
22、10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10
23、Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K1
24、0Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O34(2013 南充)计算 2(3)的结果是()A5 B1 C 1 D 5 5(2013 乐山)计算：12_.6(2013 重庆)计算：4(2)0|5|(1)2 012132.【模拟预测】1下列各数中，最小的数是()A0 B1 C 1 D2 2若|a|3，则 a 的值是()A 3 B3 C13D 3 3下列计算正确的是()A(8)80 B12(2)1 C(1)01 D|2|2 4如图，数轴上A，B 两点对应的实数分别为1 和3，若点 A 关于点 B 的对称点为C，则点 C 所表示
25、的实数是()A231 B13 C23 D231 5(1)实数12的倒数是 _(2)写出一个比4 大的负无理数_6 若 将三个数3，7，11表示在数轴上，其中能被如图所示的墨迹覆盖的数是_ 7 定 义 一 种 运 算 ，其 规则为a b1a1b，根据这个规则，计算2 3 的值是_ 8如图，物体从点A 出发，按照AB(第 1 步)C(第 2 步)DAEFGAB的顺序循环运动，则第2 012 步到达点 _处9计算：|2|(1)2 012(4)0.参考答案【考点探究】触类旁通1 C因为 5 是整数，37是分数，4 2 是整数触类旁通2 A因为 5 的相反数是 5，15的相反数是15，15的相反数是 1
26、5.触类旁通3B 触类旁通4C因为 0.055102,0.0055103,0.000 55104,0.000 055 105，故选 C.触类旁通5B因为|m3|0，且(n2)20，又因为|m3|(n2)20，所以m30文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 H
27、F2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1
28、HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1
29、 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D
30、1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2
31、D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y
32、2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2
33、Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3且 n20.所以 m3，n2，所以 m2n32(2)1 触类旁通6 A因为根据正数大于 0,0 大于负数，正数大于负数，解答即可【经典考题】1C3 131，13的倒数是 3.2CA 中，|2|2，是正数，故本选项错误；B 中，(2)24，是正数，故本选项错误；C 中，2 0，是负数，故本选项正确；D 中，2242，是正数，故本选项错误3C因为科学记数法的形式为a10n，用科学记数法表示较大的数，其规律为1a10，n 是比原数的整数位数小1 的正整数，
34、所以60 110 000 000 6.0111010.4A原式 23 5.512根据负数的绝对值是它的相反数，得1212.6解：原式 215 198.【模拟预测】1D因为正数和0 都大于负数，21，两个负数比较大小，绝对值大的反而小，所以2最小2D绝对值为3 的数有 3 和3 两个，且互为相反数3B(8)816，12(2)1，(1)01，|2|2.4A因为数轴上A，B 两点对应的实数分别为1 和3，所以 OA1，OB3.所以 ABOBOA31.由题意可知，BC AB31.所以 OC OBBC3(31)231.5(1)2(2)42(答案不唯一)67因为 30，113,173.756因为 2312
35、13362656.8A由题意知，每隔8 步物体到达同一点，因为2 012 8251 余 4，所以第2 012 步到达 A 点9解：原式 211 2.文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4
36、T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G
37、4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10
38、G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL1
39、0G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL
40、10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:C
41、L10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3文档编码:CL10G4T2Y2D1 HF2N3K9K10Z10 ZC6Z4K5J2O3