2022年2022年绝对值的意义及应用 .pdf

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1、relationship, establishe d equivalent relationship 14, and subject: appli cation problem (4)-scores and percentage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units 1) (2) find associat

2、e volume rate corresponds to relationship, Then in-line sol ution. Category fraction multiplication word problem score Division applications engineeri ng problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of com mon units of measu

3、rement and their significa nce i n rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly use d time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method o

4、f relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)-line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angl e of classification (slightly) 17, and subject: geometry pre

5、liminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column i

6、s divided into: cuboi d, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surfa ce area and volume 1, size 2, table .和绝对值的意义及应用绝对值是初中代数中的一个重要概念，应用较为广泛 在解与绝对值有关的问题时，首先必须弄清绝对值的意义和性质。对于数x 而言，它的绝对值表示为：|x|. 一. 绝对值的实质：正实数与零的绝对

7、值是其自身，负实数的绝对值是它的相反数，即也就是说， |x| 表示数轴上坐标为x 的点与原点的距离。总之，任何实数的绝对值是一个非负数，即|x| 0，请牢牢记住这一点。二. 绝对值的几何意义：一个数的绝对值就是数轴上表示这个数的点到原点的距离。例 1. 有理数 a、b、c 在数轴上的位置如图所示，则式子|a|+|b|+|a+b|+|b-c|化简结果为( ) A2a+3b-c B3b-c Cb+c Dc-b ( 第二届“希望杯”数学邀请赛初一试题) 解： 由图形可知a 0，cb0，且 |c| |b| |a| ，则 a+b 0，b-c 0所以原式 -a+b+a+b-b+c b+c，故应选 (C)

8、三. 绝对值的性质：1. 有理数的绝对值是一个非负数，即|x| 0，绝对值最小的数是零。2. 任何有理数都有唯一的绝对值，并且任何一个有理数都不大于它的绝对值，即x|x| 。3. 已知一个数的绝对值，那么它所对应的是两个互为相反数的数。4. 若两个数的绝对值相等，则这两个数不一定相等( 显然如 |6| |-6|，但 6-6) ，只有这两个数同号，且这两个数的绝对值相等时，这两个数才相等。四. 含绝对值问题的有效处理方法1. 运用绝对值概念。即根据题设条件或隐含条件，确定绝对值里代数式的正负，再利用绝对值定义去掉绝对值的符号进行运算。名师资料总结 - - -精品资料欢迎下载 - - - - -

9、- - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 1 页，共 7 页 - - - - - - - - - relationship, establishe d equivalent relationship 14, and subject: appli cation problem (4)-scores and percentage application problem review content overview answers scores, and percentage application problem of key is: ac

10、cording to meaning, (1) determine standard volume (units 1) (2) find associate volume rate corresponds to relationship, Then in-line sol ution. Category fraction multiplication word problem score Division applications engineeri ng problem problem XV, a subject: review of the measurement of the amoun

11、t of capacity, measurement and units of measurement of common units of measurement and their significa nce i n rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly use d time units and their relationships. (Slightly) with a measurement units Zhijian of of

12、 poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)-line and angle review content line, and segment, and Ray, and vertical, and parallel,

13、 and angle angl e of classification (slightly) 17, and subject: geometry preliminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid

14、 content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboi d, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surfa ce area and volume 1, size 2, table .和例 2. 已知： |x-2|+x-20，求： (1)

15、x+2的最大值； (2)6-x的最小值。解： |x-2|+x-20， |x-2|-(x-2) 根据绝对值的概念，一个数的绝对值等于它的相反数时，这个数为负数或零，x-2 0，即 x2，这表示 x 的最大值为2 (1) 当 x2 时， x+2 得最大值 2+24；(2) 当 x2 时， 6-x 得最小值 6-2 4 2. 用绝对值为零时的值分段讨论即对于含绝对值代数式的字母没有条件限制或限制不确切的，就需先求零点，再分区间定性质，最后去掉绝对值符号。例 3. 已知 |x-2|+x与 x-2+|x|互为相反数，求x 的最大值解： 由题意得 (|x-2|+x)+(x-2+|x|)0，整理得 |x-2

16、|+|x|+2x-20 令|x-2|0，得 x 2，令 |x| 0，得 x0 以 0，2 为分界点，分为三段讨论：(1)x 2 时，原方程化为x-2+x+2x-2 0，解得 x1，因不在x2 的范围内，舍去。(2)0 x2 时，原方程化为2-x+x+2x-2 0，解得 x0 (3)x 0 时，原方程化为2-x-x+2x-20，从而得x0 综合 (1) 、(2) 、(3) 知 x0，所以 x 的最大值为0 3. 整体参与运算过程即整体配凑，借用已知条件确定绝对值里代数式的正负，再用绝对值定义去掉绝对值符号进行运算。例 4. 若|a-2|2-a，求 a 的取值范围。解：根据已知条件等式的结构特征，

17、我们把a-2 看作一个整体，那么原式变形为|a-2|-(a-2)，又由绝对值概念知a-20，故 a 的取值范围是a2 4. 运用绝对值的几何意义即通过观察图形确定绝对值里代数式的正负，再用绝对值定义去掉绝对值的符号进行运算例 5. 求满足关系式 |x-3|-|x+1|4 的 x 的取值范围解： 原式可化为 |x-3|-|x-(-1)|4 它表示在数轴上点x 到点 3 的距离与到点 -1 的距离的差为4 由图可知，小于等于-1 的范围内的x 的所有值都满足这一要求。名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 -

18、- - - - - - 第 2 页，共 7 页 - - - - - - - - - relationship, establishe d equivalent relationship 14, and subject: appli cation problem (4)-scores and percentage application problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine s

19、tandard volume (units 1) (2) find associate volume rate corresponds to relationship, Then in -line sol ution. Category fraction multiplication word problem score Division applications engineeri ng problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and uni

20、ts of measurement of common units of measurement and their significa nce i n rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly use d time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly

21、 method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)-line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angl e of classificatio

22、n (slightly) 17, and subject: geometry preliminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are d

23、ivided into: cylinder and cone 2, column is divided into: cuboi d, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surfa ce area and volume 1, size 2, table .和所以原式的解为x -1 五 . 有关绝对值知识的应用1. 如果根据已知条件或题目中的隐含条件可以确定绝对值符号内的数(

24、或代数式 ) 为“负”值或“非负”值，则由绝对值的定义可直接写出其结果. 例 6. 设 x，y， a是实数，并且 |x| 1-a ，|y| (1-a)(a-1-a2)，试求 |x|+y+ a2+1的值等于_解：显然 |x| 0，|y| 0，由 |x| 0 得 1-a 0，由 |y| 0得 1-a 0，1-a 0，从而 x 0，y0，a 1 原式 |0|+0+12+12 2. 如果根据已知或题目自身不能确定绝对值符号内的代数式为“负”或“非负”，就应分别对各种情况进行讨论。讨论的方法有：(1) 直接利用绝对值的性质，去掉绝对值符号，把式子转化为不含绝对值的式子进行讨论。例 7. 已知 |a| 3

25、， |b| 2，求 a+b 的值。解： |a| 3，|b| 2， a 3 或-3 ，b2 或-2 因此 a，b 的取值应分四种情况：a3，b2 或 a3， b-2 或 a-3 ，b2 或 a-3，b-2 ，从而易求 a+b 的值分别为 5，1，-1，-5 解这类问题，要正确组合，全面思考，谨防漏解。 (2) 采用零点分区间法，求出绝对值的零点，把数轴分成相应的几个区间进行讨论( 所谓绝对值的零点就是使绝对值符号内的代数式等于零的字母所取值在数轴上所对应的点) 。例 8. 化简： |1-3x|+|1+2x|解：由031x和021x得两个零点：31x和21x，这两个点把数轴分成三名师资料总结 -

26、- -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 3 页，共 7 页 - - - - - - - - - relationship, establishe d equivalent relationship 14, and subject: appli cation problem (4)-scores and percentage application problem review content overview answers scores, and percentage application

27、 problem of key is: according to meaning, (1) determine standard volume (units 1) (2) find associate volume rate corresponds to relationship, Then in-line sol ution. Category fraction multiplication word problem score Division applications engineeri ng problem problem XV, a subject: review of the me

28、asurement of the amount of capacity, measurement and units of measurement of common units of measurement and their significa nce i n rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly use d time units and their relationships. (Slightly) with a measureme

29、nt units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)-line and angle review content line, and segment, and Ray, and v

30、ertical, and parallel, and angle angl e of classification (slightly) 17, and subject: geometry preliminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axisymmetric graphics perimeter and area combination graphics of area subject : Preliminary knowled

31、ge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboi d, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is slightly solid surfa ce area and volume 1, size 2, table .和部分：（

32、1）当21x时，031x，021x原式;5)21()31(xxx（2）当3121x时，031x，021x原式xxx2)21 ()31 (；（3）当31x时，031x，021x，原式 -(1-3x)+(1+2x)5x3. 利用绝对值的几何意义解含绝对值的方程，这样既直观，又简便。因为 |x| 的几何意义是表示数轴上点x 到原点的距离， 因此 |x-a|的几何意义是表示点 x到点 a 的距离由此可知，方程 |x-a|k 的解是 x a+k 或 x a-k(k 0) 例 9. |x-2|+|x-1|+|x-3|的最小值是 ( ) A1 B2 C3 D4 解： 设 A(1) ，B(2) ，C(3) ，

33、P(x) ，如图所示，求|x-1|+|x-2|+|x-3|的最小值，即是在数轴上求一点P，使 AP+BP+PC 为最小，显然，当P与 B重合，即 x2 时，其和有最小值2，故应选 (B) 4. 利用“一个实数的绝对值是一个非负数”这一性质解题，可使问题化难为易。在运用这一性质时，常与非负数的性质：“有限个非负数的和为零时，则每一个非负数必为零”联用。例 10. 若|m+1|+|2n+1|0，那么 m2003-n4_六. 绝对值化简与求值的基本方法例 11. 若 a、b 互为相反数， cd 互为负倒数则|a+b+cd| _(96 年泰州市初中数学竞赛 ) 解：由题设知a+b 0，cd-1 ，则

34、|a+b+cd| |0-1|1 例 12. 若|x-y+2|与 |x+y-1|互为相反数，则xy 的负倒数是 _(95 年希望杯邀请赛初一培训题 ) 名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 4 页，共 7 页 - - - - - - - - - relationship, establishe d equivalent relationship 14, and subject: appli cation problem (4)-scores and percentage ap

35、plication problem review content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units 1) (2) find associate volume rate corresponds to relationship, Then in -line sol ution. Category fraction multiplication word problem sco

36、re Division applications engineeri ng problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their significa nce i n rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitt

37、ed) 2, commonly use d time units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry prel

38、iminary knowledge (1)-line and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angl e of classification (slightly) 17, and subject: geometry preliminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axisymmetric g

39、raphics perimeter and area combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboi d, square 3, cone cone of the features of cuboids and cubes relationship between characteristic

40、s of circular cone is slightly solid surfa ce area and volume 1, size 2, table .和解： 由题设知 |x-y+2| 0，|x+y-1|0，但二者互为相反数，故只能x-y+2 0，x+y-10 解得21x，23y，43xy其负倒数是34例 13. 已知 a、b 是互为相反数， c、d 是互为负倒数， x 的绝对值等于它的相反数的2 倍，则 x3+abcdx+a-bcd 的值是 _(94 年希望杯邀请赛初一试题) 解： 由题设知 a+b 0，cd-1 又 x 的绝对值等于它的相反数的2 倍，x0，原式 03+0+a-b

41、(-1) a+b0 例 14. 化简 |x+1|+|x-2| 令 x +1 0，x-2 0，得 x-1 与 x2，故可分段定正负再去符号(1) 当 x-1 时，原式 -(x+1)-(x-2)-2x+1 ；(2) 当-1 x2 时，原式 (x+1)-(x-2)3；(3) 当 x2 时，原式 x+1+(x-2)2x-1 说明 ：例 14 中没有给定字母任何条件，这种问题应先求零点，然后分区间定正负再去绝对值符号，这种方法可归纳为：“求零点，分区间，定性质，去符号”。例 15. 设 x 是实数， y|x-1|+|x+1|。下列四个结论：.y 没有最小值；. 只有一个 x 使 y 取到最小值；. 有有

42、限多个x( 不只一个 ) 使 y 取到最小值；. 有无穷多个x 使 y 取到最小值。其中正确的是 ( )名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 5 页，共 7 页 - - - - - - - - - relationship, establishe d equivalent relationship 14, and subject: appli cation problem (4)-scores and percentage application problem revie

43、w content overview answers scores, and percentage application problem of key is: according to meaning, (1) determine standard volume (units 1) (2) find associate volume rate corresponds to relationship, Then in-line sol ution. Category fraction multiplication word problem score Division applications

44、 engineeri ng problem problem XV, a subject: review of the measurement of the amount of capacity, measurement and units of measurement of common units of measurement and their significa nce i n rate 1, currency, length, area, volume, unit size, volume, weight and rate. (Omitted) 2, commonly use d ti

45、me units and their relationships. (Slightly) with a measurement units Zhijian of of poly 1, and of method 2, and poly method 3, and of method and poly method of relationship measurement distance of method 1, and tool measurement 2, and estimates 16, and subject: geometry preliminary knowledge (1)-li

46、ne and angle review content line, and segment, and Ray, and vertical, and parallel, and angle angl e of classification (slightly) 17, and subject: geometry preliminary knowledge (2)-plane graphi cs review conte nt triangle, and edges shaped, a nd round, and fan axisymmetric graphics perimeter and ar

47、ea combination graphics of area subject : Preliminary knowledge (3)-review of solid content category 1-d shapes are divided into: cylinder and cone 2, column is divided into: cuboi d, square 3, cone cone of the features of cuboids and cubes relationship between characteristics of circular cone is sl

48、ightly solid surfa ce area and volume 1, size 2, table .和A B C D(1993 年全国初中数学竞赛试题) 解：原问题可转化为求x 取哪些值时，数轴上点x 到点 1 与点 -1 的距离之和为最小。从数轴上可知，区间-1 ，1 上的任一点x 到点 1 与点 -1 的距离之和均为2；区间 -1 ，1 之外的点x 到点 1 与点 -1 的距离之和均大于2，所以函数y|x-1|+|x+1|当-1 x1时，取得最小值2，故选 (D) 七. 绝对值与非负数我们称不是负数的有理数为非负有理数，简称非负数。 当我们说 x 是一个非负数时，用数学符号表示

49、就是x0. 值得注意的是，有的同学们往往用x0 表示任意一个非负数，而忘掉等号！这是因为他们错将非负数理解为负数的相反数了！尽管只是丢掉一个零，在数轴上只差一个点，但就全体有理数而言，却是丢掉了三类有理数中的一类。也就是说， |x| 表示数轴上坐标为x 的点与原点的距离。我们看到，任何有理数的绝对值都是一个非负数，而任何一个非负数都可表示为某数的绝对值。即对任意有理数x 有|x|0，这一点至关重要。只有牢牢掌握绝对值总是非负数并且清楚地认识到什么是非负数，才会正确地处理各种问题。例 16. 若 a 为任意实数，则下列式子中一定成立的是( )A|a| 0 B|a| a C. aa1 D. 01a

50、对这个问题的分析首先要注意到绝对值都是非负数，而非负数包括零。如此就很容易淘汰掉 A、B，而 C需从 a 的取值范围来讨论，如21a，则 C不对，至于D有非负数的性质：“一个非负数加上一个正数，得正数”，即可知其正确。例 17. 已知 a 0c，ab0，|b| |c| |a| ，化简 |a+c|+|b+c|-|a-b|解：分析这个题目的关键是确定a+c、b+c、a-b 的符号，根据已知可在数轴上标出a、b、c 的大致位置，如图所示：名师资料总结 - - -精品资料欢迎下载 - - - - - - - - - - - - - - - - - - 名师精心整理 - - - - - - - 第 6