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1、1.2.11分数的简便计算1.2.11.1变形巧算例1(1)×37=(1-)×37=1×37-×37=37-=36(2)27×=(26+1)×=26×+=15+=15(3)73×=(72+)×=72×+×=9+=9(4)×27+×41=×9+×41=×(9+41)=×50=30(5)×+×+×(6)原式=×+×+×原式==(++)×==×=1=(7)166÷41(8)1998÷1998原式=(164+2)÷41原式=1998÷=164÷41+÷41=1998÷=4+=1998×=4=例2、分析:因为有带分数存在,再观察各分式分母特征,可将带
2、分数整数部分和分数部分分开,分别求和。解:原式=(2006-1+2-3+4-…+2004-2005)+(…+)=【(2-1)+(4-3)+…+(2006-2005)】+()×1003这可让他犯了难,施工现场距离项目部很远,没有车还真是不方便office,branchoffices(jurisdiction),riskmanagement,marketingmanagementsectorthroughsupervisionandinspectionfoundproblems,shouldbeassignedtheinvestigatorsa
3、recorrectedinatimelymanner.27ththefifthchapterpenaltyunderanyofthefollowingacts,thentherelevantprovisionstopunishtheinvestigators,accordingtotheBank.Toconstituteacrimeshallbeinvestigatedforcriminalresponsibility:(A)onthebusinessthatarenotinvolvedintheinvestigation,issuedas
4、urvey.(B)customercreditinformationarenotthoroughverification.(Iii)toparticipateincreditcustomersurveyisnotinplace,customersanddataisincomplete,untrue;heknowsbearacounterfeitedclientsissuingcredit.(D)doesnotprovideforduediligenceofcreditbusiness,pre-loaninvestigationform,co
5、ncealingfactsorprovidingfalseinformationorshouldbefoundinanormalinvestigationfailedtodiscovertheriskfactors,leadtothereviewandapprovalpolicyerrors,loanrisk.(Five)onloanguaranteesofsurveynotinplace,notbyprovidesonarrived,andpledgerealforfieldverification,andassessment,andid
6、entificationandregistration,notaccordingtoprovidesonguarantorofguaranteesqualificationandguaranteescapacityforsurveyverified,ledtoguaranteesloanlostauthenticity,andlegitimacy,andeffectivenessof;cycleloanbusinessintheofmortgage=1003+167=1170例3、分析:此题直接计算太麻烦了,通过观察,发现从第三个分数开始,
7、往后数到,这8个分数的计算结果正好是0,如果从再往后数8个数,其结果也是0,那么从开始到止,中间有2002-3+1=2000个分数,每8个一组,正好250组。因为这250组每组计算结果都是0,因此有如下简单解法。解:原式=1++=11.2.11.2拆分法(也叫裂项法)这可让他犯了难,施工现场距离项目部很远,没有车还真是不方便office,branchoffices(jurisdiction),riskmanagement,marketingmanagementsectorthroughsupervisionandinspectionfoun
8、dproblems,shouldbeassignedtheinvestigatorsarecorrectedinatimelymanner.27ththefifthchapter