《优秀资料(2021-2022年收藏)小学奥数04分数简便计算.doc》由会员分享,可在线阅读,更多相关《优秀资料(2021-2022年收藏)小学奥数04分数简便计算.doc(7页珍藏版)》请在taowenge.com淘文阁网|工程机械CAD图纸|机械工程制图|CAD装配图下载|SolidWorks_CaTia_CAD_UG_PROE_设计图分享下载上搜索。
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 13737 37 36 (2) 27(26+1)26+15+15 (3)73(72+)72+9+9 (4)27+419+41(9+41)50 30(5)+ (6)原式+ 原式(+) 1(7)16641 (8) 19981998原式(164+2)41 原式199816441+41 19984+ 19984 例2、分析:因为有带分数存在,再观察各分式分母特征,可将带分数整数部分和分数部分分开,分别求和。解:原式(20061+23+4+20042005)+(+)【(21)+(43)+(20062005)】+()1003100
2、3+1671170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,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+12000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)例9:(+)(+)1例11:解:设S 那么3S1+ 得3S-S12S则S例12:1+1+2()1+2()11.2.11.3分数运算的其他杂题 例1:解:设x,y,z原式(x+y)(y+z)(x+y+z)yxy+y2+xz+yzxyy2yzxz1例2:看下
3、面几个算式:找出上面三个最简真分数求和的规律。再算一下下面两题:(1)(2)求分母是1001的最简真分数的和等于多少。分析:仔细观察数列规律,不难发现:由此可以得出最简真分数求和规律:有几个最简真分数,和就是几除以2。解:(1)1226(2)分母是100171113 那么,分子就不能是7或11或13的倍数。从11000中,7的倍数有143个,11的倍数有90个,13的倍数有76个。这其中同时是7和11倍数的有12个,同时是7和13的倍数的有10个,同时是11和13倍数的有6个,同时是7、11、13倍数的有0个。那么是7或11或13倍数的数有: 143+90+7612106+0280(个) 则满
4、足题意的有:1000280720(个),那么和应该是7202360, 所以分母是1001的所有最简真分数的和为360。例3:依题意知道分母没变,而与的分母不同。由此可知这两个分数已约分。约前的分母只能是4、10的公倍数。假定分母是20,则这两个分数应是、,分子差5-2=3。原分数的分子加3、减3后所得的两个分数的分子应差6.显然这两个分数的分子和分母应同乘以2(63),即、。所以原分数为,或。例5 将的分子加上某数,分母减去同一个数,则分数约分后变为,求这个数。解:因为新分数约分后为,可设分子为3份,分母为5份。又知分子分母的和没变,可得:1份(17+55)(3+5)7289,分子为3927,
5、分母为5945,变化后的分数为。271710,所以那个数为10。例6 把的分子和分母同时加上什么数,可以得到?因为一个分数的分子、分母同时加上一个数,分子与分母的差是不变的。由17314,532,1427(倍),知新分数已用7约分。故未约分时的分数是 这个“什么数”是:351718或21318。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, branch offices (jurisdiction), risk management, marketing management sector through supervision and inspection foun
6、d problems, should be assigned the investigators are corrected in a timely manner. 27th the fifth chapter penalty under any of the following acts, then the relevant provisions to punish the investigators, according to the Bank. To constitute a crime shall be investigated for criminal responsibility:
7、 (A) on the business that are not involved in the investigation, issued a survey. (B) customer credit information are not thorough verification. (Iii) to participate in credit customer survey is not in place, customers and data is incomplete, untrue; he knows bear a counterfeited clients issuing cre
8、dit. (D) does not provide for due diligence of credit business, pre-loan investigation form, concealing facts or providing false information or should be found in a normal investigation failed to discover the risk factors, lead to the review and approval policy errors, loan risk. (Five) on loan guar
9、antees of survey not in place, not by provides on arrived, and pledge real for field verification, and assessment, and identification and registration, not according to provides on guarantor of guarantees qualification and guarantees capacity for survey verified, led to guarantees loan lost authenticity, and legitimacy, and effectiveness of; cycle loan business in the of mortgage