预览加载中,请您耐心等待几秒...
1/9
2/9
3/9
4/9
5/9
6/9
7/9
8/9
9/9

在线预览结束,喜欢就下载吧,查找使用更方便

如果您无法下载资料,请参考说明:

1、部分资料下载需要金币,请确保您的账户上有足够的金币

2、已购买过的文档,再次下载不重复扣费

3、资料包下载后请先用软件解压,在使用对应软件打开

relationship,establishedequivalentrelationship14,andsubject:applicationproblem(4)--scoresandpercentageapplicationproblemreviewcontentoverviewanswersscores,andpercentageapplicationproblemofkeyis:accordingtomeaning,(1)determinestandardvolume(units"1")(2)findassociate"volumeratecorrespondsto"relationship,Thenin-linesolution.CategoryfractionmultiplicationwordproblemscoreDivisionapplicationsengineeringproblemproblemXV,asubject:reviewofthemeasurementoftheamountofcapacity,measurementandunitsofmeasurementofcommonunitsofmeasurementandtheirsignificanceinrate1,currency,length,area,volume,unitsize,volume,weightandrate.(Omitted)2,commonlyusedtimeunitsandtheirrelationships.(Slightly)withameasurementunitsZhijianofofpoly1,andofmethod2,andpolymethod3,andofmethodandpolymethodofrelationshipmeasurementdistanceofmethod1,andtoolmeasurement2,andestimates16,andsubject:geometrypreliminaryknowledge(1)--lineandanglereviewcontentline,andsegment,andRay,andvertical,andparallel,andangleangleofclassification(slightly)17,andsubject:geometrypreliminaryknowledge(2)--planegraphicsreviewcontenttriangle,andedgesshaped,andround,andfanaxisymmetricgraphicsperimeterandareacombinationgraphicsofareasubject:Preliminaryknowledge(3)-reviewofsolidcontentcategory1-dshapesaredividedinto:cylinderandcone2,columnisdividedinto:cuboid,square3,coneconeofthefeaturesofcuboidsandcubesrelationshipbetweencharacteristicsofcircularconeisslightlysolidsurfaceareaandvolume1,size2,table...和函数解析式的七种求法一、待定系数法:在已知函数解析式的构造时,可用待定系数法.例1设f(x)是一次函数,且f[f(x)]4x3,求f(x).建议收藏下载本文,以便随时学习!解:设f(x)axb(a0),则f[f(x)]af(x)ba(axb)ba2xabb2a2a2a4,.或abb3b1b3f(x)2x1或f(x)2x3.二、配凑法:已知复合函数f[g(x)]的表达式,求f(x)的解析式,f[g(x)]的表达式容易配成g(x)的运算形式时,常用配凑法.但要注意所求函数f(x)的定义域不是原复合函数的定义域,而是g(x)的值域.11例2已知f(x)x2(x0),求f(x)的解析式.xx2111解:f(x)(x)22,x2,f(x)x22(x2).xxx三、换元法:已知复合函数f[g(x)]的表达式时,还可以用换元法求f(x)的解析式.与配凑法一样,要注意所换元的定义域的变化.例3已知f(x1)x2x,求f(x1).解:令tx1,则t1,x(t1)2.f(x1)x2x,f(t)(t1)22(t1)t21,f(x)x21(x1),f(x1