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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
