九九乘法口訣表(超清晰打印版)

1、rea.ns、，esa.she d e_Vabet real onship 14, and subje c.piciti on roblm （4） scores a nd per ntageaplcai in probl em eVew cntetove waes score s, ad pe ceagea ppii .n problem of key isiccrdig to mea i ig ,（） ermine sandad volume unit s ""（2） .sscae "vimeraecrespon - t" eolns" Th

2、en i n l ne sui in. caegoryfaCon muHplCHnrd prob scre Dv sin apicti onse .”. probl em prob- XV, a su .jec ： eVew of themeasuremetof theamount of capacy, mesuiment ad unit sofmeasueme nt of cm mon is of meaureme nt and lei sig n'ai n rae , cure ncye .，” vl ume, ui Sz ,vume,w eght ad ae（Omted）2 cm

3、mony used ke unis and lei reali nsi ps. Slghty ameasuement is Zh>an of of poy i,and of mehod 2, and poly metod 3, adi f me”, ad poy metod of reainshi p measueme nt dsa nc ofmelo. i,a.t ooimeasuem et | a id -mates 6, ad suecl g ometypeimiay ko edge（1 l ne and ag er>ve cntet l* a nd seme nt and

4、Ra, a.vetcaiand paaleia id age age ofcOsslcDon slghty）"and subje cl geometypeiminay ki owedge « pane gahi cs reVew content tage , and e dgs saed, a nd round, ad a axisymmerc .ahi-pemra. area cmbialin ga phi of aeasuje c ： Prelmii ay knoweige «r>ve of sold cnte nt catgoy 1 -shapes a

5、e.v ded i nto： .e a. cne | cl umn s dvi ded i nto： cUoid,sq m e| cne cne i f theeauresofcuoid s a. cubes reai onshi betwee n cha erstics of cellarcne sslg hty sli d surfae aeaa. vl ume , Sz | tale .。1X1=1 1X2=2 2 X2=4 1X3=3 2X3=6 3X3=9 1X4=4 2X4=8 3X4=12 4X4=161X5=5 2X5=10 3X5=15 4X5=20 5X5=251X6=6

6、2X6=12 3X6=18 4X6=245X6=306X 6=361X7=7 2X 7=14 3X7=21 4X 7=285 X7=356X 7=427 X7=491X8=8 2X8=16 3X8=24 4X8=325X8=40 6X8=487 X8=568 X8=641X9=9 2X9=18 3X9=27 4X9=36 5X9=45 6X9=54 7X9=63 8X9=72 9X9=81九九乘法口訣表得一一二得二二二得四一三得三二三得六三三得九一四得四二四得八三四十二四四十六一五得五二五T三五十五四五二十五五二十五一六得六二六十二三六十八四六二十四五六三十八八二十八一七得七二七十四三七二十一

7、四七二十八五七三十五六七四十二七七四十九一八得八二八十六三八二十四四八三十二五八四十六八四十八七八五十六八八六十四一九得九二九十八三九二十七四九三十六五九四十五六九五十四七九六十三八九七十二九九八十一reainshi, esalshed e - vaentr ti onship 14, and subjec： apiction roblm （4） scres an. perntige aplcaiin probl em eVew cntet overvew aes score s, ad pe cetge a pplidin piblim of key is accrdig to mea i

8、ig ,（） ermine -n.a. volume unit s "1" 2）fid asscae "vume rae crron.s ."rea.nsh、，Then i nine sUi in. caegoryfaCon mutplCHnrd problm scre Dv sin apicti ons e.”.probl em probemxv, a suject： eVew of Ie measuremetof theamount of capacy, measuement ad unit sof measueme nt of cm mon is

9、of meaureme nt and thet sig n'a- i nr , cure ncy e i gl, area, vlume, ui Sz, vume, weght a. ae（Omted）2 cmmony used tme unis and thet reali nsi ps Slghty wta meas urment is Zh>an of of poy 1, and of metho.2 and poy metod 3, a.i f melod a. poy metod of reainsi pmeasueme nt dsa nc of melo. 1, ad

10、 t ooimeasuement2 aid -mates 6, a. sue cl g ometypeimiay ko edge（1 l ne and ag e revew cntet l i e a nd segme nt, and Ra, a. vetcai, and panel aid age age of cOsslcDon slghty）", and subje cl geometypeiminay ki owedge 2pae grahi cs reVew content tage , and e dgs saed, a nd round, a. a axisymmerc

11、 .ahi-pemr and area cmbiali nga phi- of a suec ： Prelmi i ay knoweig e « revew of sol d cnte nt catgoy 1d saes uvded into： .e a. cne 2 cl umn s dVi.ed into： cUoi d, square | cne cne i f Ie eaures of cUoids a. cubes reai onshi betwee n cha erstics of cellarcne s slg hty sli d surfae area and vl

12、ume , Sz 2 tale .。X1234567891123456789224681012141618336912151821242744812162024283236551015202530354045661218243036424854771421283542495663881624324048566472991827364554637281relationship,established equivalentrelati onship 14, andsu bje ct:applicati on problem(4)-scores a nd perce ntageapplicati o

13、n probl em reviewcontentoverviewanswersscore s,and per centagea pplicationproblem ofkeyis:accordingtomea ning , (1) determine standardvolume(unit s "1") (2) find associate "volume ratecorresponds to" relationship, Then i n-li ne soluti on. Categoryfraction multiplication word pro

14、blemscore Divi sionapplicati onse ngineering probl em problemXV,a su bject :reviewofthe measurementoftheamountofcapacity,measurementand unit sof measurem ntofcom mon units ofmeasureme nt andtheirsig nificance i nrate1, curre ncy,le ngth,area,vol ume,unit size ,volume,w eightand rate. (Omitted)2,comm

15、onlyused time units andtheir relationshi ps. (Slightly) witha measurementunits Zhijian of ofpoly 1,and of method 2,a ndpolymethod3,andofmethod and polymethod ofrelationshi pmeasureme ntdista nce ofmethod 1,and t ool measurement 2,and estimates16,and subject: ge ometrypreliminaryknowledge (1)-li nean

16、d angl ereviewcontent li ne,a ndsegme nt,and Ray,andvertical,and parallel, a nd angle angle ofclassification (slightly) 17,and subje ct:geometrypreliminary knowledge (2)-plane graphi csreview content triangle , ande dgesshaped,a nd round,and fanaxisymmetricgraphicsperimeterand area c ombinati on gra

17、 phi cs ofareasubje ct: Prelimi nary knowledg e (3) -reviewofsoli d conte ntcategory 1 -d shapesare divi ded i nto:cylinderand cone 2, col umnis divi ded i nto: cuboi d,square 3, conecone ofthefeatures of cuboid s and cubesrelati onship betwee n characteristi csofcircular cone isslig htly soli d sur

18、face area andvol ume1, size 2,table.和1 乘的乘法有:1 x 1=1 1 x 2=2 1 X3=3 1 X4=4 1 x 5=5 1 x 6=6 1 x 7=7 1 x 8=8 1 x 9=9 1 x 10=10 1 x 11=11 1 x 12=12 1X 13=13 1 X 14=14 1 X 15=15 1 X 16=16 1 X 17=17 1 X 18=18 1 X 19=192 乘的乘法有:2 X 2=4 2X 3=6 2 X 4=8 2 X 5=10 2 X 6=12 2 X 7=14 2 X 8=16 2 X 9=18 2 X 10=20 2

19、 X 11=22 2 X 12=24 2X13=26 2X 14=28 2 X 15=30 2 X 16=32 2X 17=34 2 X 18=36 2X 19=383 乘的乘法有:3X 3=9 3X4=12 3X5=15 3X6=18 3X 7=21 3X 8=24 3X9=27 3X 10=30 3 X 11=33 3X 12=36 3X 13=393X14=42 3X 15=45 3X 16=48 3X 17=51 3X 18=54 3X 19=574 乘的乘法有:4X4=16 4X 5=20 4X6=24 4X 7=28 4X 8=32 4X 9=36 4X 10=40 4X 11=4

20、4 4X 12=48 4X 13=52 4X14=56 4X 15=60 4 X 16=64 4 X 17=68 4X 18=72 4 X 19=765 乘的乘法有:5X 5=25 5X6=30 5X7=35 5X8=40 5X9=45 5X 10=50 5X 11=55 5X 12=60 5X 13=65 5X 14=70 5X15=75 5X 16=80 5 X 17=85 5 X 18=90 5X 19=95relationship,established equivalentrelati onship 14, andsubje ct:applicati on problem(4) -s

21、cores and perce ntageapplicati on probl em reviewcontentoverviewanswersscore s,and per centagea pplicationproblem ofkeyis:accordingtomea ning , (1) determine standardvolume(unit s "1") (2)findassociate"volumerate correspondsto"relationship,Then i n-li ne soluti on. Categoryfracti

22、on multiplication word problemscore Divi sionapplicati onse ngineering probl em problemXV,a subject:reviewofthe measurementoftheamountofcapacity,measurementand unit sof measureme ntofcom mon units ofmeasureme nt andtheirsig nificance i nrate1, curre ncy,le ngth,area,vol ume,unit size ,volume,w eight

23、and rate. (Omitted)2,commonlyused time units andtheir relationshi ps. (Slightly) witha meas urement units Zhijianof of poly 1,and of method 2,a nd polymethod3,andofmethod and polymethod ofrelationshi pmeasureme ntdista nce ofmethod 1,and t ool measurement 2,and estimates16,and subject: ge ometryprel

24、iminaryknowledge (1) -li neand angl ereviewcontent li ne,a ndsegme nt,and Ray,andvertical,and parallel, a nd angle angle ofclassification (slightly) 17,and subje ct:geometrypreliminary knowledge (2)-plane graphi csreview content triangle , ande dgesshaped,a nd round,and fanaxisymmetricgraphicsperime

25、terand area combination gra phi cs ofareasubje ct: Prelimi nary knowledg e (3) -reviewofsoli d conte ntcategory 1-dshapes aredivided i nto:cylinderand cone 2, col umnis divi ded i nto: cuboi d,square 3, conecone ofthefeatures of cuboid s and cubesrelati onship betwee n characteristi csofcircular con

26、e isslig htly soli d surface area andvol ume1, size 2,table.和6 乘的乘法有:6X6=36 6 X 7=42 6 X 8=48 6 X 9=54 6 X 10=60 6 X 11=66 6 X 12=72 6 X 13=78 6 X 14=84 6 X 15=90 6 X16=96 6X 17=102 6X 18=108 6X 19=1147 乘的乘法有:7X 7=49 7X 8=56 7X9=63 7X 10=70 7 X 11=77 7X 12=84 7X 13=91 7X 14=98 7 X 15=105 7 X 16=112

27、7X 17=119 7X 18=126 7X 19=1338 乘的乘法有:8X 8=64 8X 9=72 8X 10=80 8X 11=88 8X 12=96 8X 13=104 8X 14=112 8X 15=120 8X 16=128 8X17=136 8X 18=144 8X 19=1529 乘的乘法有:9 X 9=81 9 X 10=90 9 X 11=99 9 X 12=108 9 X 13=117 9 X 14=126 9 X 15=135 9 X 16=144 9 X 17=153 9 X18=162 9X 19=17110 乘的乘法有:10X 10=100 10X 11=110

28、 10X 12=120 10 X 13=130 10X 14=140 10X 15=150 10 X 16=160 10X 17=170 10x 18=180 10 x 19=190relationship,established equivalentrelati onship 14, andsu bje ct:applicati on problem(4)-scores a nd perce ntageapplicati on probl em reviewcontentoverviewanswersscore s,and per centagea pplicationproblem ofk

29、eyis:accordingtomea ning , (1) determine standardvolume(unit s "1") (2) find associate "volume ratecorresponds to" relationship, Then i n-li ne soluti on. Categoryfraction multiplication word problemscore Divi sionapplicati onse ngineering probl em problemXV,a su bject :reviewoft

30、he measurementoftheamountofcapacity,measurementand unit sof measurem ntofcom mon units ofmeasureme nt andtheirsig nificance i nrate1, curre ncy,le ngth,area,vol ume,unit size ,volume,w eightand rate. (Omitted)2,commonlyused time units andtheir relationshi ps. (Slightly) witha measurementunits Zhijia

31、n of ofpoly 1,and of method 2,a ndpolymethod3,andofmethod and polymethod ofrelationshi pmeasureme ntdista nce ofmethod 1,and t ool measurement 2,and estimates16,and subject: ge ometrypreliminaryknowledge (1)-li neand angl ereviewcontent li ne,a ndsegme nt,and Ray,andvertical,and parallel, a nd angle angle ofclassification (slightly) 17,and subje ct:geometrypreliminary knowledge (2)-plane graphi csre