有用的小學(xué)到初中所有數(shù)學(xué)公式大全

有用的小學(xué)到初中所有數(shù)學(xué)公式大全（Usefulmathematicsfromelementaryschooltojuniormiddleschool）1,thenumberofcopiesofeachnumberx=totalnumber,eachnumberofcopies,copiesnumber==eachnumber2,1timesofafewafewmultiplesofmultiplesofmultiplesofx=1,=severalmultiplesofmultiplesofmultiplesofmultiples/=1times3,speedtimedistance,distance=x=distance/timetimespeed=speed4,thenumberofthetotalprice,pricex===thenumberoftotalnumberofunitprice,unitprice5,workefficiency*workingtime=totalamount,totalworkefficiency=workingtimeThetotalamountofwork,worktime=workefficiency6,theaddendaddend=+and-addend=otheraddend7,theminuend-==-differencedifferencereductionminuenddifference+=minuendmeiosismeiosis8,factorXfactor=productdividedbyafactor=anotherfactor9,dividend,dividend,divisor=businesstaking=x=dividenddividenddivisorquotientMathematicalformulasforelementaryschoolmathematics1,square:Cperimeter,Sarea,alengthsidelength=perimeter=4C=4aarea=lengthside*S=a*a2,thecube:V:volumea:edgelengthsurfacearea=lengthxlengthx6S=a*a*6Volume=edgelengthxxlong*xV=axa*a3,rectangle:Cperimeter,Sarea,a,sidelength,perimeter=(length+width)*2C=2(a+b)area=**widthS=ab4,cuboidV:volumes:,areaa:long,b:width,h:height(1)surfacearea(length*width+length*height+width*height)*2S=2(ab+ah+bh)(2)volume=long*width*highV=abh5,triangleSahhighendareabottomarea=xheight/2s=ah/2Triangleareahigh=x2/bottomTrianglearea=x2/high6,parallelogram:sareaa,bottomh,higharea=bottom*highs=ah7,ladder:sareaabhhighbottombottomarea=(+downonthebottomend)*/2s=(a+b)*H/28round:SCd=r=diameterradiuscircumferencepi(1)==2***perimeterdiameterradiusC=d=2pipipi*r(2)***area=radiusradius9,cylinder:V,volumeh:,highS:bottomareaR:bottomradiusC:bottomperimeter(1)thesidearea=thecircumferenceofthebottom*high(2)surfacearea=sidearea+basearea*2(3)volume=basearea*height(4)=2/Xvolumesidearearadius10,V:hshighvolumeconebottomareaRbottomradiussize=baseareaxheight/3Thetotal,thetotalnumberofcopies=averagenumberEquationofsumanddifference(+anddifference)830192=number(andpoor)830192=decimalDoublingproblemAnd(1=decimalmultiples).Decimal=multiple=largenumber(orsum=decimal=largenumber)Differenceproblem(1/differencemultiples)=decimalDecimal=multiple=largenumber(ordecimal+difference=largenumber)Treeplantingproblem1,theproblemoftreeplantingonnonclosedlinescanbedividedintothefollowingthreecases:Ifanonclosedlineatbothendsareplantingtrees,so:Number=numberof+1/1=lengthspacing*(1=lengthspacingnumber)Fulllength(1/spacing=number)Ifyouwanttoplanttreesintheendofnonclosedline,theotherendisnotplanting,so:Number=number=length/spacingLength=xnumberofrows=length/numberofrowsIfbothendsinnonclosedlinearenotplantingtrees,so:Numbersectionnumber=-1/1=lengthspacing*(number=lengthspacing+1)=length/spacing(number1)2,thenumberoftreesonclosedlinesisasfollowsNumber=number=length/spacingLength=xnumberofrows=length/numberofrowsProfitandloss(profit+loss),thetwodistributionofthedifferencebetweenthenumberofcopiesinthedistributionof=(largesurplus-smallsurplus)/twodistributionofthedifferencebetweenthenumberofcopiesinthedistributionof=(burned-asmalllossoftwotimes),thedifferencebetweenthenumberofcopiesisassignedtoparticipateinthedistributionoftheEncounterproblemEncounterdistance=speedandXmeetingtimeMeettime=distance/speedandmeetThespeedanddistance,theencountertime=meetCatchupwithproblemsTrackingdistance=speeddifference*trackingtimeChaseandchaseanddistance/time=speeddifferenceThespeeddifferenceanddistance=chasechaseandtime.FlowproblemDownstreamvelocity=hydrostaticvelocity+flowvelocityCountercurrentvelocity=hydrostaticvelocity-flowvelocityHydrostaticspeed=(downstreamvelocityandvelocity),2Flowrate=(downstreamspeedvelocity),2ConcentrationproblemTheweightofthesolute+theweightofthesolvent=theweightofthesolutionTheweight,theweightofthesolutesolutionx100%=concentrationWeightofsolution=concentration=soluteweightTheweight,concentrationofsolutesolution=weightProfitanddiscountproblemsProfit=sellingprice-costTherateofprofit,profit=costX100%=(sold/cost1)*100%Amountoffluctuation=principal*ChangePercentageTheactualpricediscount,theoriginalpricex=100%(<1discount)Interest=principal*interestrate*timeAftertaxinterestrates***=principaltime(120%)LengthConversion1kilometers1meters=1000meters=10.1cm=100cm=101meterdecimeter1cm=10mmAreaConversion1sqkm=100hectares1hectares=10000squaremeters1squaremeters=100squaredecimeter1square=100squarecentimeterdecimeter1mm2=100mm2Volume(volume)productunitconversion1cubicmeters=1000cubicdecimeter1cubicmeter=1000cubiccentimeter1DM3=1L1cubiccentimeter=1ml.1cubicmeters=1000litersWeightunitconversion1tons=1000kg1kilograms=1000grams1kg=1kgUnitconversionofRenminbi1yuan=10Jiao1jiao=10minutes1yuan=100centstimeconversionFirstCentury=100,1years=12monthsMonth(31days):1\3\5\7\8\10\12monthXiaoyue(30days)are:4\6\9\11monthThereare28daysinFebruary,aleapyear29daysinFebruaryThereare365daysayear,aleapyear366daysayear1days=24hours,1hours=60minutes1minutes,=60seconds,1hours,=3600secondsMathematicalformulasofgeometry,perimeter,areaandvolumeofPrimaryMathematics1,theperimeteroftherectangle=(long+width)*2C=(a+b)*22,squareperimeter=sidelength*4C=4a3,rectangulararea=long*widthS=ab4,squarearea=lengthside*sidelengthS=a.a=a5,thebottomofthetrianglearea=xheight/2S=ah/26.Theareaoftheparallelogram=bottom*highS=ahArea=7,trapezoidal(bottom+bottom)xheight/2S=(a+b)H/28x2d=2rdiameter=radiusradius=2D/2r=diameter.9,thecircumferenceofacirclediameterx=pi=pi**2c=D=2radiusPiPiR10,theareaofacircleradiusradius=pi**Commonformulasforjuniorhighschoolmathematics1thereisonlyonelineaftertwo2thelinebetweentwopointsistheshortest3thesameangleorisometricequalmargin4withtheangleofthecomplementofequalorconstant5havingonlyonelineandperpendiculartotheknownline6,thelineoutsideapointandthelineonallpointsconnectedbytheline,theverticallinesegmentistheshortest7parallelaxiomsfollowapointoutsidethestraightlineandhaveonlyonelineinparallelwiththeline8ifthetwolinesareparalleltothethirdlines,thetwolinesareparalleltoeachother9correspondinganglesareequal,thetwoparallellines10alternateanglesareequal,thetwoparallellines11complementaryinterioranglesonthesameside,twoparallellines12thetwoparallellines,correspondinganglesareequal13twolineparallel,alternateanglesareequal14thetwoparallellines,complementaryinterioranglesonthesamesideThesumof15sidesofatriangleisgreaterthanthird16inferthatthedifferencebetweenthesidesofatriangleislessthanthird17triangletrianglesumtheoremthreeanglesandequalto180degrees18inference1thetwoacuteanglesofrighttriangleoverlapeachother19acorollaryofatriangleisequaltothe2cornersandtwoanglesitisadjacentand20acorollaryofatrianglewith3cornersthananyoneanditisnotadjacent.Thecorrespondingsidesof21congruenttrianglesandequalangles22edgeaxioms(SAS)arecongruentwithtwotrianglesofequalanglesonbothsidesandtheirangles23cornercorneraxiom(ASA)hastwotrianglescongruenthornsandtheiredgescorrespondingtotheequal24inference(AAS)hadtwohornsandoneoftheangleonthesidecorrespondingtotheequaltwocongruenttriangles25sideaxioms(SSS)havingtwoequaltrianglesequaltothreesides26bevelandrightsideaxioms(HL)havetwoequalanglesofrighttriangleandonerightangleCongruent27theorem1thedistancebetweenpointsonthebisectorofanangletothesidesofthisangleisequalThepointatwhichthedistancebetween2and28anglesisthesameasinthebisectorofthisangleThebisectorof29anglesisthesetofallpointsequaltothetwosidesoftheangleTwoand30areequalisoscelestriangletheoremofanisoscelestriangle(i.e.equalequilateralangle)31inferthatthebisectorofthe1isoscelestrianglevertexisequaltothebaseandperpendiculartothebase32thetopanglesoftheisoscelestriangle,thebisector,themiddlelineonthebottomedgeandtheheightonthebottomedgecoincidewitheachother33corollary3equilateraltriangleswhoseanglesareequal,andeachangleequalto60degrees34thedecisiontheoremofisoscelestriangle,ifatrianglehastwoanglesequal,thenthesetwoanglesTheoppositesidesareequal(equalangles,equalsides)35corollary1threeanglesareequaltriangleisanequilateraltriangle36corollary2anisoscelestrianglewithanangleequalto60degreesisanequilateraltriangle37inarighttriangle,ifanacuteangleisequalto30degrees,therightanglededgeofitisequaltothebevelededgehalf38thecenterlineonthebevelofarighttriangleisequaltohalfofthebevelThepointatwhichthetwoendsofthelineofthelineofdivisionareequaltothe39endsofthelineApointequaltothedistancebetweentwoendpointsof40inversetheoremandalinesegment,intheverticalbisectorofthelineTheverticalbisectorofa41linesegmentcanbeconsideredasasetofallpointsequaltothedistancebetweenpointsattheendofalinesegment42theorem1thetwographsofastraightlinesymmetryarecongruentforms43theorem2iftwographsaresymmetricalaboutastraightline,thenthesymmetryaxisisperpendiculartothelineofthecorrespondingpointBisector44theorems3,twographsaresymmetricalaboutaline,iftheircorrespondinglinesorextensionlinesintersect,ThentheintersectionpointisonthesymmetryaxisThe45inversetheorem,ifthecorrespondinglinesofthetwographsareverticallydividedbythesameline,thentwoAgraphrelatingtothesymmetryofaline46Pythagoreantheoremofrighttriangletworightanglesidesa,bsquare,andisequaltothesquareofthehypotenusec,Thatis,a^2+b^2=c^247thePythagoreantheoremandinversetheoremofa,ifthethreesideoftriangleB,Ca^2+b^2=c^2,ThenthetriangleisrightangledtriangleIntheorem48quadrilateralandequalto360degrees49quadrilateralcornersandisequalto360degreesInthe50polygonsumtheoremnedgeshapeandisequalto(n-2)*180degrees51anyinferencemultilateralangleandequalto360degreesThepropertytheoremof52parallelogram;thediagonalequalityof1parallelogramThepropertytheoremof53parallelogram;theequalsidesof2parallelogram54parallelsbetweentwoparallellinesareequalThepropertytheoremof55parallelogram;diagonalbisectorof3parallelogram56parallelogramdecisiontheorem1.Thetwogroupsofequalanglesareparallelogram57parallelogramdecisiontheorem2.ThequadrilateraloftwosetsofequalsidesisparallelogramThe58parallelogramtheorem;3diagonalsequaltooneanother;thequadrilateralisaparallelogram59theparallelogramtheorem4isaparallelogramofparallelpairsofedgesequaltoaparallelogramFouranglesof60rectangulartheorem1rectangleisright61rectangularpropertytheorem2thediagonalofarectangleisequal62rectangulardecisiontheorem1hasthreeanglesatrightangles,andaquadrilateralisarectangle63rectangulardecisiontheorem2aparallelogramwithequaldiagonalsisarectangle64diamondtheorem1thefoursidesofadiamondareequal65diamondtheorem2thediagonalsofthediamondareperpendiculartoeachotherandeachdiagonallineisdividedintoasetofdiagonalanglesHalfofthe66diamondarea=diagonalproduct,namelyS=(a*b)/267diamonddecisiontheorem1quadrilateralisequaltofoursidesisdiamond68diamonddecisiontheorem2thediagonalsareperpendiculartoeachotherandtheparallelogramisrhombicThe69squaretheoremfouranglesof1squareisarightangle.Thefoursidesareequal70thesquaretheoremof2squaresequaltotwodiagonalsandperpendiculartoeachotherAdiagonalsetofdiagonallinesThe71theorem1,aboutthecentralsymmetryofthetwographsarecongruent72theorem2relatingtothecentralsymmetryofthetwofigures,thesymmetrypoints,theconnectionisthroughthecenterofsymmetry,andbeSymmetriccentralbisectionThe73inversetheorem,ifthecorrespondingpointsofthetwographsareconnectedatacertainpointandequaltothispoint,Well,thesetwographsaresymmetricalaboutthispoint74isoscelestrapezoidpropertytheoremtheisoscelestrapezoidisequaltotwoanglesatthesamebottom75isoscelestrapeziumwithtwodiagonalsequal76isoscelestrapezoidjudgmenttheorem,onthesamebottom,thetwoanglesequaltrapezoidisisoscelestrapezoid77thetrapeziumwithequaldiagonalsisisoscelestrapezoidAtheoremof78parallellines,ifthelinesegmentsofasetofparallellinesareequalinastraightline,Thenthelinesegmentthatiscutontheotherlineisequal79inference1,aftertheladder,themiddleofawaistandthebottomparallelline,willbeequaltotheotherwaist80inference2thelinebetweenthemidpointofthetriangleandtheothersideisequaltothethirdside81themedianlineofatriangleinwhichthemedianlineofatriangleisparalleltothethirdsideandequaltohalfofitThemedianlinetheoremofthe82laddertheorem.ThemedianlineofaladderisparalleltothebottomofthetwoandequaltohalfofthesumofthetwobasesL=(a+b)/2S=L*h83(1)thebasicnatureoftheratio,ifa:b=c:d,thenad=bc,ifad=bc,thena:b=c:d84(2),iftheratioofa/b=c/D,then(a+b)/b=(c+D)/D85(3)geometricproperties,ifa/b=c/d=...=m/N(b+d+...+n(a+c+=0),then...+m)(b+d+...+n)=a/b86parallellinesaredividedintolinesegments,andthreeparallellinesarecutintotwostraightlinesProportion87inferenceparalleltothetriangularsideoftheline,cuttingtheothersides(ortheextensionlineonbothsides)ProportionaltolinesegmentsThe88theorem,ifalinecutsthesidesofatriangle(ortheextensionlinesonbothsides),thecorrespondinglinesThelineisparalleltothethirdsideofthetriangleAtriangleof89paralleltoonesideofatriangleandintersectingwiththeothersidesThethreesideisproportionaltothethreesidesoftheoriginaltriangle90theoremsparalleltoonesideofatriangle,intersectingwiththeothersides(orextendedlinesonbothsides),Thetriangleformedissimilartotheoriginaltriangle91similartrianglestheorem1cornersequaltwotriangles(ASA)92rightangledtrianglesaredividedbytheheightoftwosidesofabevel.Thetriangleissimilartotheoriginalone93decisiontheorem2,thetwosidesareproportionaltoeachotherandtheanglesareequal,andthetwotrianglesaresimilar(SAS)94decisiontheorem3,threesidescorrespondingproportional,twotrianglesimilarity(SSS)The95theorem,iftherightangleofarighttriangleandtherightanglededgeofanotherrighttriangleThebevelisproportionaltoarightangle,sothetworighttrianglesaresimilar96propertytheorem1theratioofthecorrespondingtriangles,theratioofthecorrespondinglinetothebisectorofthecorrespondingangleTheratioisequaltotheratio97propertytheorem2theratiooftheperimeterofasimilartriangleisequaltotheratioofthesimilarity98propertytheorem3theratiooftheareaofasimilartriangleisequaltothesquareofthesimilarityratio99sinerandomacutevalueequaltoitsanglecosinevalue,anyacuteanglecosinevalueequaltoitsCoanglesinevalueTangent100anyacuteanglevalueequaltoitscomplementofthecotangentvalue,arbitrarycotangentvalueequaltoitsangleThetangentangleThe101circleisthesetofpointswhosedistanceisequaltothefixedlengthTheinnerpartofa102circlecanberegardedasasetofpointswhosedistanceissmallerthantheradiusTheouterpartofa103circlecanberegardedasasetofpointswhosedistanceisgreaterthantheradiusWith104orsoisequaltotheradiusofthecircleThedistancefrom105tothepointisequaltothelocusofthefixedpoint.ThecircleisfixedasthecenterofthecircleandtheradiusisfixedThetrajectoryofapointequaltothedistancebetweentwoendpointsofaknownlinesegment,theverticalbisectorofalinesegment106Thetrajectoryofapointequaltothedistancebetween107sidesofaknownangle;thebisectorofthisangleThetrajectoryofpointsfrom108totwoparallellinesthatareparallelandequidistantfromthetwoparallellinesAstraightlineThe109theoremdoesnotdetermineacircleatthreepointsonthesameline.110theverticaldiametertheoremisperpendiculartothediameterofthechord,equaltothestring,andthetwoarcsofthestringareequallydivided111inference1Thediameterofthechord(notthediameter)isperpendiculartothestringandthetwoarcsofthestringareequallydividedTheverticalbisectorofthechordpassesthroughthecenterofthecircleanddividesthetwoarcsofthestringequallyEqualtothediameterofanarctowhichthechordisright,verticallyequaltothechord,andequaltotheotherarcofthestring112inferthatthearcofthetwoparallelstringsofthe2circleisequalThe113circleisacentralsymmetricalfigurecenteredonthecenterofthecircle114theoremsinthesamecircleorcongruentcircles,arcequalcentralangleof,thechordsareequal,ThechorddistancesofthepairsofstringsareequalInthe115roundorrounddeduction,ifthetwocentralangle,twoarcs,twostringsortwostringstringThereisasetofequalamountsinthecenterofthepitch,andtherestofthemareequalThe116theorem,thecircumferenceofapairofarcsisequaltohalftheangleofthecenterofthecircleitiscenteredon117corollary1witharcorarconthecirclewithequalangles;orcircle,equalcircumferentialangleThearcsareequal118inferthatthe2circle(ordiameter)istherightangleofthecircle;thecircleof90degreesistherightchordDiameter119inference3ifthecenterlineononesideofthetriangleequalshalfoftheside,thenthetriangleis...RighttriangleComplementarydiagonalinscribedquadrilateral120circletheorem,andanyoneofthecornersareequaltoithorn121linesLandOintersect,d<RTheLandO/d=rtangentlineThelinearLandOfromD/>R122thejudgmenttheoremoftangentsfollowstheouteredgeoftheradiusandthestraightlineperpendiculartotheradiusisthecirclecutLine123thenatureofthetangenttheorem;thetangentofthecircleisperpendiculartotheradiusofthetangentpoint124inference1thestraightlinepassingthroughthecenterofthecircleandperpendiculartothetangentmustpassthroughthepointoftangency125inference2thestraightlinepassingthroughthepointoftangencyandperpendiculartothetangentmustpassthroughthecenterofthecircle126thelongtangenttheorem,thetwotangentsthatdrawthecirclefromapointoutsidethecircle,theirtangentsareequal,thecenterofthecircleandTheconnectionofthispointdividestheanglebetweenthetwotangentsThetwosetsof127sidesofacircumscribedquadrilateralandequalThe128angleisequaltothecircumferenceoftheclampingangletheoremofarcangle129thatifthetwoanglebetweenthearcareequal,thenthetwoangleisequalThetwointersectingstringinthecircleof130intersectingchordtheorem,theproductofthetwolinesdividedbytheintersectionpointisequaltotheproductofthesamelength131itisdeducedthatifthechordisperpendiculartothediameter,thenhalfofthestringisthetwolineofitsdiameterTheratioofmean132thecuttinglinetheoremfollowsthetangentandsecantofacirclefromapointoutsidethecircle.ThetangentlengthisthepointtotheSecantandthecircleThetwolineslongtermofproportionintersection133thetwosecantofacirclefromapointoutsidethecircle,whichistwooftheintersectionofeachSecantandthecircleTheproductofalinesegmentisequal134Ifthetwocirclesaretangent,thenthepointoftangencymustbeonthecenterline135,two,D,R+r,two,d=R+r,two,D,R-r,R+r(R>R)Twocircleinscribedd=R-r(R>R).Thetwocirclecontainsd<R-r(R>R)Acommonchordinwhich136linesoftwointersectingcirclesareverticallydividedintotwoequalcirclesDividethecircleinto137theoremsofn(n=3):Theconnectingpointsofthepolygoninturnisthecircleinscribedregularnpolygon.Afterallthepointsaretangenttoacircle,theintersectionadjacentvertextangentpolygonisthecircleThecircumscribedpositivenedge138anytheorempolygonhasacircumscribedcircleandacircle,thesetwocirclesareconcentricEachcorner139regularnpolygonisequalto(n-2)*180degrees/nThe140theoremistheradiusandtheedgedistanceofthepositivenedge.Thepositivenedgeisdividedinto2ncongruentrighttriangle141,theareaofthepositivenedgeisSn=pnrn/2P,whichrepresentstheperimeterofthepositivenedgeThe142isthetrianglearearoot3A/4asidesaid143IfthereisanangleofKpositivenedgesaroundavertex,thesumoftheseanglesshouldbe360degreesduetoThisk*(n-2)180deg/n=360isconvertedto(n-2)(K-2)=4The144arclengthformula:L=nwasR/180The145sectorareaformula:Ssector=nWuR^2/360=LR/2146insidelength,=d-(R-r),grandfathertangentlength=d-(R+r)Utilities:commonmathematicalformulasFormula,formula,formula,expressionMultiplicationandfactorization,a2-b2=(a+b)(a-b),a3+b3=(a+b)(a2-ab+b2)A3-b3=(a-b(a2+ab+b2))Thetriangleinequalityislessthanorequalto|a|+|b|lessthanorequalto|a|+|b||a-b||a+b||a|=b<=>-b=a=b|a-b|=|a|-|b|-|a|=a=|a|-b+havethesolutionsofquadraticequationwithoneunknownofthe(b2-4ac)/2a-b-(b2-4ac)/2aVRelationbetweenrootsandcoefficientsX1+X2=-b/a,X1*X2=c/anotes:thelawsofWeberDiscriminantB2-4ac=0note:equat