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绝对值的意义及应用

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2021-01-25 03:49
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谚语大全-

2021年1月25日发(作者:康恺)
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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,

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|.

.
绝对值的实质:

正实数与零的绝对值是其自身,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。


.
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。


1. 有理数
a

b

c
在数轴上的位置如图所示,则式子< br>|a|+|b|+|a+b|+|b-c|
化简结果

( )

A

2a+3b-c B

3b-c C

b+c D

c-b
(
第二届“希望杯”数学邀请赛初一试题
)
解:
由图形可知
a

0

c

b

0
,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)



.
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。


.
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。

relationship, establi
she
d equivalent relati
onship
14, and subje
ct: a
ppli
cation
problem (4)--score
s and percentage a
ppli
cation
problem review conte
nt overview a
nswer
s scores, a
nd perce
ntage applicati
on pr
eliminary knowle
oblem
of key is: accordi
ng to meani
phi
ng, (1) determine sta
ndar
d vol
ume (units
n in-line
sol
ution. Category fracti
on multi
plicati
on w
ord pr
oblem score
Division applicati
ons engi
neeri
ng problem problem XV, a subje
ct: review
of the measur
ement of the am
ount
of capa
city, measurement a
nd
units
of measurement of com
nt and their sig
nifica
nce i
n rate 1, currency, le
ngth, area, v
olume,
unit si
ze, vol
ume, weight a
nd rate. (Omitted) 2, commonly
use
d time unit
s and t
heir relationships. (Slightly) with a measureme
nt units Zhijia
n of of poly 1, and of method 2, a
nd
poly method 3, a
nd
of method and
poly method
of relationshi
p measureme
nt dista
nce
of method
1, and tool mea
sureme
nt 2, and estimates 16,
and subject: ge
ometry preliminary k
nowle
dge (1)--li
ne a
nd a
ngle review conte
nt line, a
nd segment, and Ray, a
nd vertical
, and parallel, a
nd a
ngle angl
e of classification (slightly) 17, and subje
ct: geometry pr
dge (2)--plane gra
cs review
conte
nt triangle, a
nd e
dges shape
d, a
nd round, and fan axisymmetric graphi
cs perimeter and are
he
a com
binati
on graphics of area subje
ct : Preliminary knowledge (3)-review
of soli
d content categ
ory 1-d shape
s are divided int
o: cyli
nder a
nd
cone 2, colum
n is divi
ded i
nto:
cuboi
d, square 3,
cone cone
of the features of cuboids a
nd
cubes relati
mon units of measureme
onship
betwee
n characteristi
cs of cir
cular
cone is slig
htly soli
d surfa
ce area and vol
ume 1, size 2, table ...


2.
已知:
|x-2|+x-2

0


求:
(1)x+2
的最大值;
(2)6-x
的最小值。
< br>解:

|x-2|+x-2

0
,∴
|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
x

2
,这表示
x< br>的最大值为
2
(1)

x

2
时,
x+2
得最大值
2+2

4


(2)

x

2
时,
6-x
得最小值
6-2
=< br>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。


3.
已知
|x-2|+x

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|+|x|+2x-2
0

|x-2|

0
,得
x
=< br>2
,令
|x|

0
,得
x

0

0

2
为分界点,分为三段讨论:

(1)x< br>≥
2
时,原方程化为
x-2+x+2x-2

0
,解 得
x

1
,因不在
x

2
的范围内,舍去 。

(2)0

x

2
时,原方程化为
2 -x+x+2x-2

0
,解得
x

0
(3)x

0
时,原方程化为
2-x-x+2x-2

0
, 从而得
x

0
综合
(1)

(2)
、< br>(3)

x

0
,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。


4.

|a-2|

2-a
,求
a
的取值范围。

解< br>:根据已知条件等式的结构特征,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-2

0
,故
a
的取值范围是
a

2
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.


5.
求满足关系式
|x-3|-|x+1|< br>=
4

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。

relationship, establi
she
d equivalent relati
onship
14, and subje
ct: a
ppli
cation
problem (4)--score
s and percentage a
ppli
cation
problem review conte
nt overview a
nswer
s scores, a
nd perce
ntage applicati
on pr
eliminary knowle
oblem
of key is: accordi
ng to meani
phi
ng, (1) determine sta
ndar
d vol
ume (units
1
n in
-line
sol
ution. Category fracti
on multi
plicati
on w
ord pr
oblem score
Division applicati
ons engi
neeri
ng problem problem XV, a subje
ct: review
of the measur
ement of the am
ount
of capa
city, measurement a
nd
units
of measurement of com
nt and their sig
nifica
nce i
n rate 1, currency, le
ngth, area, v
ume 1, size 2, table ...
olume,
unit si
ze, vol
ume, weight a
nd rate. (Omitted) 2, commonly
use
d time unit
s and t
heir relationships. (Slightly) with a measureme
nt units Zhijia
n of of poly 1, and of method 2, a
nd
poly method 3, a
nd
of method and
poly method
of relationshi
p measureme
nt dista
nce
of method
1, and tool mea
sureme
nt 2, and estimates 16,
and subject: ge
ometry preliminary k
nowle
dge (1)--li
ne a
nd a
ngle review conte
nt line, a
nd segment, and Ray, a
nd vertical
, and parallel, a
nd a
ngle angl
e of classification (slightly) 17, and subje
ct: geometry pr
dge (2)--plane gra
cs review
conte
nt triangle, a
nd e
dges shape
d, a
nd round, and fan axisymmetric graphi
cs perimeter and are
he
a com
binati
on graphics of area subje
ct : Preliminary knowledge (3)-review
of soli
d content categ
ory 1-d shape
s are divided int
o: cyli
nder a
nd
cone 2, colum
n is divi
ded i
nto:
cuboi
d, square 3,
cone cone
of the features of cuboids a
nd
cubes relati
mon units of measureme
onship
betwee
n characteristi
cs of cir
cular
cone is slig
htly soli
d surfa
ce area and vol


所以原式的解为
x

-1

.
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
(
或代数式
)

“负”值或“非负”值,则由绝 对值的定义可直接写出其结果
.

6.

x

y

a
是实数,并且
|x|

1-a

| y|

(1-a)(a-1-a
2
)
,试求
|x|+y+< br>a
2
+1
的值等

______



:显然
|x|

0

|y|

0


∴由
|x|

0

1-a
0
,由
|y|

0

1-a
0



1-a

0
,从而
x

0

y

0

a

1
∴原式=
|0|+0+1
2
+1

2
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。


7.
已知
|a|

3

|b|

2
,求
a+b< br>的值。

解:

|a|

3

|b |

2



a

3

-3

b

2

-2
因此
a

b
的取值应分四种情况:

a

3

b

2

a

3

b

-2

a

-3

b

2

a

-3

b

-2< br>,

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
(

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
)



8.
化简:
|1-3x|+|1+2x|



解:

1
?
3
x
?
0

1
?
2
x
?
0
得两个零点:
x
?
1
1

x
?
?

这两个点把数轴分成三
3
2

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