中考一轮复习讲义第1章有理数(含答案)

1、Coose the corec meani ng；（4）tcrec the ypos； （5） so the ci d wre w ords aba B, and :.A bb;（6） i acor da - wl h w.e .word*（ 7 the cmpiee wr d, and explain t he meani ng oft he wo”8）c.i on；（9） make sentences wl h the wo（10） the We lag uage as require -（C t he main seen - tyes（） cmpbehe meai ng of a see

2、 nne or - pe on f thought s ad ig s 3 W - I e nces - reLilie d； 4 iismigment of the - nenne；（5 moidkd -ne I ies 2, knowlld.cssiicai on （1）the cmmon cnjunCins cordiate .I. , to exmie the t opi c dei，pr obems Isciao wi h tw 2, anlysisalerave in tws i die ct pr oporlon t o Ie - ount of I e asoc atd el

3、ainsi p s inverey prporl onalreainshi p. | and letink now, col nproporinye 4, ad sluions prporl on tye 5 a nd e s, w ote aser an|_ge pe nay and lubjec：appli catap proem a nd com piste aplCat on probem evew cntent simpl e apical compose a ppl cat emaes a geea sep* ad fgue oU mea nig througg mils t he

4、, Ind know n cndiins a nd by se ekng pr oem 2, ad a -yss numbe reat oship a- ss kow n cndiins Z hija n, and cndtons and proem Zh o eaos .eemi ne pr oem solving me hod ad pr oem solving s es. 3, a ndclumn y pe ua.nlss f omua, i s oU subdivi Sons 4, ad teS, ad woeanswececk, ad ceki ng a nd w oe answer

5、 s typical appl cat a nd suec： appliai prb3)cl umn e .ainsl u.ns a cntet oveViw prbemsv I g ps 1, ad fgue oU mea nig, I nd by seek ng of unk now n a nd x | ad a _rdig tmeai ng Ind eq u ntrinship, l ss E qua! on | ad sut I ns eq - .n 4, a nd teS, a nd woe a nsw esiccrding tm-ning fid e qUva et reaton

6、si of cmmon mehod 1 , And acrdig to cmm reaionsiy pe - a blishd-Uv nt eat I nsi p | adiccrdig to has lar “hd of d culaion Im u| and a it he of focus desci be d sentene lom . al Shag detemie bascof -uv nt eat onsi|4, ad usig 一me ntue ad lit method, mehod a numbe中考数学一轮复习精品讲义第一章有理数知识网络结构图整数按定义分厂一分类管正包婪

7、按符号分回一（盍有理相反数;只有符号不同的两个数互为相反数伍嬴赢积屯丽两个数互为 赢绝对值:数轴上表示数甘的点与原点i至.的绝对值是其本身,的距离？做值的绝对值Jo的绝对值是0II负蝙一对酒良它的相:数卜科学记数法:把一个绝对值大型。的数表示成”1/的形式（1应时10有效数字;从一个数的左边第一个不为0的数字起，到末位数字止，有 理 数:所有的数字都悬这个数的有效数字同号两数相加，取相同的符号，绝对值相加j （鼻号两数相加：取绝次值大的符号，重超值相减/，互为相反数的两个数相加为0H数同。相加仍得这个数减法;减去一个数，等于加上这个数的相反数法则，:后目得定;绝好值相乘（ 卜乘法异号得负:绝对

8、值相I任高薪百。相乘；都得。I.渚：除以一个不等于0的数，等于乘这个数的倒数 正数的任何次嘉都是正数：奇次落是负数偶次舞是正数10侬任何正整数次瀛是0加法交换律4交换律工二:-/一义乘法交攒律一智法加法结合律；口+（匕+同=佃+与+CI玛算律结合律y：乘法结合律:岫c=a（加）=（ab）c分配律 ab+c)=abacobem a nd com posteap.lon prbem eviw cnte nt simpl e apl-t composea aswes a of geea seps, ad Igiue oiU m- ning througg eamins t he, Ind kow n

9、 cnditions and by se ekng pr ob 2, ad a .a numbe reat onsip aay sis kow n cnditions Z hija n, and cndiins a nd probem Zhija of elainsi, deemi ne pr obem sving me hod ad pr obem soig s es I a nd col imn y pe ccuain lss formila, is oiU ”dV sons 4, ad tes, ad woeanswe ceckad ceki ng a nd wote anr s tyi

10、ca appl cat 3and ”ec： aplicaion prbI） col umn e .ainsl uins a cntet ovevew prbem sV I g se ps 1, ad ue oit mea nig, I nd by ei ng of unk now n a nd x | ad a ccrdig tmeai ng fnd iquvaent reainshi, lss E Liat on | ad sit I ns iqiain 4 a nd tes, and wo. a nsw esiccrding tm-ni ng fid eLil-aet reatonsi o

11、f cmmon mehod 1 ,Ad acrdig to cmmumbe reaionsiy pe a blishdIquv nt eat I nsi p | ad iccrdig to has lar .hhd of cl .laion i. ul, | a nd a it he of focis desci be se ntece fomove al Shag deemine bbscofiquvaent eatonsi|4, ad usig sgme nt ue ad lit method, mehod aayss numbe重点题型总结及应用题型一绝对值理解绝对值的意义及性质是难点，

12、由于|a|表示的是表示数a的点到原点的距离，因此ai内0可运用|a|的非负性进行求解或判断某些字母的取值.例1如果a与3互为相反数，那么|a +2|等于()A. 5 B. 1 C. 1 D. 5解析：a与3互为相反数，则 a=3,所以|a+2|= |-3+2|= |-1|= 1.答案：B例 2 若(a-1)2+|b+2|=0,则 a+ b=.解析：由于(a1)2q |b+2|却又(a1)2与|b+2|互为相反数，因此(a 1)2 = 0且|b+2|=0,则 a= 1, b= 2,所以 a +b= 1.答案：1规律若几个非负数的和为 0,则这几个数分别为 0.题型二有理数的运算有理数的运算包括加

13、减法、乘除法及乘方，是初中数学运算的基础.要熟记法则，灵活运算，进行混合运算时，还要注意运算顺序及运算律的应用.例3 (1)2 011的相反数是()A. 1 B. 1 C. 2 011 D. - 2 011解析：由于指数2 011为奇数，所以(1)2 011 = 1,其相反数为1.答案：A12)1 1例 4 计算：(1) -1 1 黑|+2 x(-8)-9 1-11;I 4八 5八，12)（2） 1 - 1 -0.5M3-2-（-3） 2 ppioblem a nd com poste apicalcoooe le core c meai ng；to crrec the yPos;(5) so

14、 the cid e w ords ABA B andA FB,(6) i accr dace wtwie I wrdsthe cmplee wr d, and -paihe m-ni ng of t he word8) clcai on；(9) make setecs wih the word(10) the wtelag uuge r iqued (C)h e main seenne typs (| cmplehe meai ng of a see nne or - pe on of thoughts ad being s wie -I e nces re qured； 4 Inis ml

15、gnment of the nenne；(5 modla -ne 一 2, knowl Idge fiai o(1)thecmmon cn.nCins cordiae -.一,to eiethe topi c deiyprobems sciaId wt h tw 2, aayssaleia twswsidi diectp oporlon to le -ount of the assc.td elainsip i nverey plporlonal realonshi p. | and letunknow n.ol n proporlon ye 4, ad sluions prporl on t

16、ye 5 a nd e s, w ote aser annu.e pe nay and lubjec：n p.bem evewcnte ntsmpl e .plcal composeappl -aes a ofgeea sep*.d *ueoU m- nig througgeamins t he, Ind kow n cnditions a nd by se ekng pr ob 2, ad a .a numbe realonsip aay ss kow n cndiins Z h n, and cndiins aprobem Zh of eainsi,deemi ne pr obem slV

17、ng me hod ad probem soig ses 3, a ndcol-n y pe in lss formi s ou lubdiv sons 4, ad tes, ad woeanswe ce CeCkad ceki ngg a nd w ote anr styica applcalonppobllm and lubec： aplicaion prblm 3) clumn e .ainsl ulons a pplcalonppobl m eVewcntet ovevewprbemsli I g se ps 1, ad fgueoU meanig, I nd by ei ng of

18、unk now Ia nd x | . a_rdng i meaing Ind iquvaent realonship, l ss E Lial on | ad sulI ns iq in 4 a nd tes, and woe a nsw esICcrdi ng lm-ni ng find equva et real onsipof cmmon mehod 1 , . acrdig to cmm r ionsiy pe - a blishdiquvaent eal I nsi p I ad iccrdng lhas lar . hhd of culaion i. u | a nd a -di

19、ng he of focus iesjibe d se nlene fomoveal Shag d .ieb of iquvaent ealon. 4, ad usng 一ment ague adi. melod, mehod aayss numbe,1解：(1) -1-(-8)-9- 11-1 2 (-8)-945=4-9X-=4 4= 0.11 -0.5 13 2-(-3) 23-9)5、=1- - m(-7)16(-7)=题型运用运算律简化运算过程运用加法的交换律、结合律，把某些具有相同属性的数(如正数、负数、分数中的分母具有倍数关系、相反数等)分别结合在一起相加，可以简化运算过程.例5

20、计算下列各题.(1)21-49. 5+10. 2-2-3. 5+19;(2)1 Q一 一2 一20 一7-32 ；3(-0.2)-2-111- 2- -13- 2442434一2g 19119 43252分析：混合运算，应按法则进行，同时注意灵活运用运算律，简化运算过程.解：(1)原式=(21+19)+10 . 2+( -49. 53. 5)-2 = 50. 2-55=- 4. 8;(2)原式一一2 - 2 -3二一一2 一-coooe Ie core c meai ng；to crrec the yPos;(5) so the cid e w ords ABA B andA FB,(6) i

21、 a_r dace wt h wie I wrdsthe cmplee wr d, and -paihe mlaning of t he word8) clcai on；(9) make setecs wih the word(10) the wtelag uuge - r (que . (C)he main seenne typs (, cmpehe meai ng of a see nne or - pess on f thoughts ad feig s 一一I e nces - re quie d； 4 is m(gment of the nence； (5 modid -ne I n

22、e s 2, knowl Idge fiai o(1)thecmmon cnjunCins cordiae -.一,to eiethe topi c deiyprobems -sciaId wt h tw 2, aayssaleraie -Isin tws i diectpr oporlon to Ie -ount of Ie asoc atd el ainsip i nve y plpolonalrinshi p. | and letunk now, col n proporlon ye 4, ad sui0ns. PrPortPPiobem a nd com poste aplcaln p

23、.bem evewcnte ntsmpl e .ptclUlemcompose app - p bl -aes a ofgeea sep*.d g-eoU m- nig througgeamins t he, Ind kow n cnditions a nd by se ekng pr ob 2, ad a .a numbe realonsip aay ss kow n cndiins Z h n, and cndiins aprobem Zh of eainsi,deemi ne pr obem slVng me hod ad probem soig es 3, a ndcol-n y pe

24、 in lss formua, i sou lubdiv sons 4, ad tes, ad woeanswe ceckad ceki ngg a nd w ote anr styica applcalonppobl a nd lubec： aplicaion p.b3col umn e .ainsl ulons a pplcalonppobl - eVewcntet ovevewprbemsli I g se ps 1, ad Igue ou mea ning, I nd by ng of unk now Ia nd x | . a_rdng i meaing Ind iq-vaent r

25、- Ioss Equal on | ad sulI ns iq in 4 a nd tes, and woe a nsw esiccrdi ng Im-ni ng find equva et real onsiof cmmon mehod 1 , . acronsiy pe - a blishdiquvaent eat I nsi p I ad iccrdng Ihas lar . hhd of culaion i. u | a nd a -ding he of focus descibe se nIene fomoveal Shag emine bbscof iq-vaent ellonsi

26、l 4, ad usng sgment fgu. ad lis melod, mehod aayss numbe13.11(3)原式2445 7 55+ +-1654 3 445+ 4044024-c7 c55 c_24 - 24241251270 56 -330 +125 = 121 =120403940(4)原式= I M 12，J_21 W 4 243 51927943 19 16825 19 43 2527八八-0=0.1点拨正、负数分别结合相加;(2)分数中，同分母或分母有倍数关系的分数结合相加;除法转化为乘法，正向应用乘法分配律;(4)逆向应用分配律 a(b+c)= ab+ac,即

27、ab+ac=a(b+c).题型四利用特殊规律解有关分数的计算题根据题目特点,灵活将算式变形，对不同算式采取运算顺序重新组合、因数分解、裂项等不同的方法，达到优化解题过程、简化计算、解决问题的目的.例6计算下列各题.5 -2 3 J(1) 5 9 +17 3;63423 17315(2) k5959 +59I;5 2 12777一2 6 12 20 30 42 56 72 9011111(4)2+16512 1 0242 048分析：(1)带分数相加，可将带分数中整数部分与分数部分拆开分别相加.(3) 本题若按常规计算方法比较麻烦，但若用运算律可简化运算.(3)由于-=2 1 23,12Coos

28、e meaning；(4) to cr-c the yci d ew o-Ban.：(6) iacorda- h wie .word*he cmpete or-8) clcai on；.” h the wor-(10 the w.e l-g uuge ty- I nsi p | ad ncr dig t has bar .hhd of iaion f. ula a nd a it he of fo-s iesci be d sentene fomove al Shag .ee.ie baic of lquv nt ei onsi| 4, ad usng 一me nt Igu. ad lit m

29、etod, m.hod aayss numbe111111111111111120=4-5=4-5， 30 - 56 5 6 ? 42 6171 6 7 56 78 7 8 ?11111111- J 一= =, 一 = -,所以将原算式变形裂项后，再进行计算.72 8 9 8 990 9 10 9 10(4)算式中，后一个分数的分母是前一个分数分母的2倍，可在算式中加上最后一个分,11,1、,人 一广, 一，人、，数，再减去，加上的与前一个分数运算，所得的和再与前一个分数运2 0482 0482 048算，依次向前进行，最终求得运算结果.5 一 2 一 3 一 1解：(1)原式=_ 59_+1

30、7 +3 6342523111=(-5- 9 + 17-3) Q| = 0-1 = _1；6342443 1(2) I -5 273155959- 59127773+7H59制3 17315 59 + -59-59 + 5 2 127773_1.Z5 2 12(59-59+59)59+1317=60- - 60- 60 = 36- 30- 35 = -295212(3)原式111111111+ +12233445566778899 10.1-11-11-11,22 33 4419_10 10(4)原式=1111+ +- + +2 4 8 16111+ ! + ! + !1024 2048 20

31、4811111.=十r i 十20482 4 8 1611111111r r - = r r r “ r r 1024 1024 20482 4 8 16512 2048Coose the corec meani ng；4tcrecthe ypos；（5） so the ci d e w ords（B- B, and BB）（6） i acor.a - wl h w.e .word*（ the cmpiee wr d, and explain t he meani ng of the wor_8.1i on；（9） make sentences wl h the wo（10） the writ

32、e lag uage as require -（C）t I e main seen - tyes（） cmpbee sentences；（2 w.e down themeaingofa see n or ex pe on of thought s a. fee. * 3 wie se i e nces as re quie d；（4 .sm-lgnment of the se ntence；（5 modfed sente . s 2, knowledge cosslcati on （1 the cmmon cnjuncins cordiate . . H. , to examine the t

33、 opi c detiy pr obems assciat ed wl h two 2, aayss, aierave -es.n tws i die ct p oporlon t o Ie amount of the assoc aed el ainshi p s inversey prporl onal reainshi p. | and set unknow n.olum n proporin ye 4, a. suionsprporlontyeIandes,woteanswera nguage plenary and sub， ec：apli apia.nprobemandcompos

34、ieap.at on prbem evew cnte nt smpl e apl composea aswes a of geea seps,ad*-e ou m- nig througg eamins t he, Ind kow n cndiins a nd by se ekng pr ob 2, ad a I ays numbe reat onsip a- ss kow n cndiins Z h n, and cndiins a nd probem Zhja of elainsi, deemi ne pr obem svng me hod ad pr obem soVig s es I

35、a nd col umn y pe ccua.n -s f ormua, i sou dil sons 4, ad tes, ad woeanswececkadchekingandwoteanr s tyica appl cat 3and”ec： aplicai on prb.) col umn e .ainsl uins a cntetovevwprbemsli I g se ps 1, ad ue out mea nig, I nd by - ei ng of unk now n a nd x | ad a ccrdig tmeai ng Ind iquv nt r eainshi, l

36、ss E quat on | ad sut I ns lq - .n 4 a nd tes, a nd woe a nsw esiccrdi ng tm-ni ng find e quva et reatonsi of cmmon mehod 1 ,Ad accrding to cmm reaionsiy pe - a blishd Iquv nt eat I nsi p | ad iccr dig to has lar .hhd of culaion i. ul, | a nd a it he of focus desci be se ntece fomove al Shag emine b

37、bscof iquv nt eat onsi| 4, ad usig 一me nt u. ad lit method, mehod aayss numbe111,12 047=+ =1 -=.2 2 2 0482 0482 048点拨利用规律特点，灵活解分数计算题，需要认真观察，注意经常训练，提高思维的灵活性.题型五有理数运算的应用用正负数可以表示相反意义的量，有理数的运算在生活中的应用十分广泛，其中，有理数的加法、减法及乘法运用较多.做题时，要认真分析，列出算式，并准确计算.例7有8箱橘子，以每箱15千克为标准，超过的千克数记为正数，不足的千克数记 为负数，现记录如下(单位：千克)：1. 2

38、, -0. 8, 2. 3, 1. 7, -1. 5, -2. 7, 2, -0. 2, 则这8箱橘子的总重量是多少？分析：本题运用有理数的加法、乘法解决问题.先求出总增减量，再求出8箱橘子的总标准重量，两者之和便为这8箱橘子的实际总重量.解析：1. 2+(0. 8)+2. 3+1. 7+( 1. 5)+( 2. 7)+2+( 0. 2)=1. 2-0. 8+2. 3+1 . 71. 5-2. 7+20. 2= (2. 3+1 . 7+2)+(-0, 8- 2. 7 1. 5)+(1 . 2-0. 2)=6 5+1 = 2.则 15X8+2= 122(千克).答案：这8箱橘子的总重量是122千

39、克.例8 一货车为一家摩托车配件批发部送货，先向南走了8千米，到达 华能”修理部，又向北走了 3. 5千米，到达 捷达”修理部，继续向北走了 7. 5千米，到达 志远”修理部， 最后又回到批发部.(1)以批发部为原点，以向南方向为正方向，用 1个单位长度表示1千米，你能够在数 轴上表示出 华能“捷达“志远”三家修理部的位置吗？(2)志远”修理部距 捷达”修理部多远？(3)货车一共行驶了多少千米？解：能.如图161所示.志远捷达 华能卜 k L E _ ! Il J.I H 1 I .-3 -2 -I n I 234567H*IO图 1-6-1(2)由数轴可知 志远”修理部距 捷达”修理部4.

40、5-(-3)=4. 5+3 = 7. 5(千米).apli apiainprobemandcomposie ap.H on prbem evew cnte nt smpl e apl compose a ppl aswes a of geea seps, ad ,ue ou m- nig througg eamins t he, Ind kow n cndiins a nd by se ekng pr ob 2, ad a ，a numbe reat onsip a- ss kow n cndiins Z hn, and cndiins a nd probem Zhja of elainsi,

41、 deemi ne pr obem svng me hod ad pr obem soVig s es I a nd col umn y pe ccua.n-s f ormua, i sou lubdv sons 4, ad tes, ad woeanswececk ad cheki ng a nd w ote anr s tyica appl cat 3 a nd “ec： aplicai on prb.）cl umn e .ainsl uins a cntet ovevw prbemsli I g se ps 1, ad ue out mea nig, I nd by ng of unk

42、now n a nd x | ad a ccrdig tmeai ng Ind iquv nt r eainshi, l ss E quat on | ad sut I ns lq - .n 4 a nd tes, a nd woe a nsw esiccrdi ng tm-ni ng fid e quva et reatonsi of cmmon mehod 1 ,Ad accrdig to cmm reaionsiy pe - a blishd Iquv nt eat I nsi p | ad iccr dig t has lar .hhd of laion i. ul, | a nd a

43、 it he of focus desci be d se ntece fomove al Shag d .ie b of iquv nt eat onsi| 4, ad usig 一me nt u. ad lit metod, mehod aayss numbe（3）货车共行驶了 |8|+|3. 5|+| 7. 5|+|3|=8+3. 5+7. 5+3= 22（千米）.题型六探索数字规律找数字规律的题目成为近几年中考的热点问题，这类题目灵活多变.解题时要认真观察、分析思考，找出规律，并运用规律解决问题.例9某种细菌在繁殖过程中，每半小时分裂一次，由一个分裂成两个，2. 5小时后，这种细菌可分

44、裂为（）A. 8 个B. 16 个 C. 32 个 D. 64 个解析：本题数字的规律是1 一2一4一8，每半小时细菌个数变为原来的2倍，所以经过2. 5小时，细菌个数应变为原来的25倍，即32个.答案：C例10 观察图162,寻找规律，在“?处应填上白数字是（）- - - 01010A. 128B. 136C. 162D. 188解析：观察图个数字特点可发现：8= 4+2+2; 14=8+4+2;26=14+8+4;.所以 88+48+26 = 162.答案：C思想方法归纳本章中所体现的数学思想方法主要有：1 .数形结合思想：在本章中，自始至终利用数轴来定义或描述有理数的概念和运算， 数轴成

45、为理解有理数及其运算的重要工具.这种把数与形（图形或数轴）结合起来进行研究的思想方法，是学习数学的重要思想方法.2 .分类讨论思想：a与-a哪个大呢？a的绝对值等于什么？在本章中，我们都是通过分 类讨论解决问题，分类讨论可以把一个复杂的问题分成若干个较简单的问题来处理，这是数学中处理问题的一种重要思想方法.不重复、不遗漏是对分类讨论提出的基本要求.例如， 我们常把有理数分成正有理数、负有理数和零三类，如果遗漏了零，只考虑正有理数和负有理数两种情况，就会犯错误.3 .转化思想：有理数的加法是通过符号法则转化为绝对值（小学所学的数）的加减法进apli apia.nprobem a nd com p

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