计算:252m÷ 题目和参考答案——精英家教网——
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计算：252m÷
答案：解析：
　　解：原式＝(52)2m÷(5－1)1－2m
　　＝54m÷52m－1
　　＝54m－(2m－1)
　　＝52m+1
　　分析：被除式底数是25．除式底数是，可把它们都化成5，再去运用同底数幂的除法性质进行计算．
　　点拨：逆运用同底数幂除法性质及幂的其他运算性质是解此类题的技巧性方法．
练习册系列答案
科目：初中数学
来源：黄冈学霸　七年级数学　下 新课标版
计算：252m÷
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请输入手机号& 列竖式计算，带★的要验算．96÷30=252÷80=★235
本题难度：0.60&&题型：解答题
列竖式计算，带★的要验算．96÷30=252÷80=★235÷76=416÷32=423÷31=★682÷34=
来源：学年新人教版四年级（上）期中数学试卷（4） | 【考点】整数的除法及应用．
列竖式计算，带※的要验算．458÷6=637÷9=405÷4=※542÷6=600÷5=※423÷8=※325÷2=607÷6=
列竖式计算（带※的要验算，带△的得数保留两位小数）．4.06×5※81.6÷0.3486.3×4.2△3.92÷3.7．
列竖式计算（带*的要验算）．*305×16=263×94=718×50=89×502=
（2016春o赣州校级月考）列竖式计算，带※的要验算458÷6=405÷4=600÷5=※325÷2=
（2014秋o梓潼县校级期中）列竖式计算，带★的要验算423+349=301-84=★409+394=900-461=646+263=★708-379=
解析与答案
（揭秘难题真相，上）
习题“列竖式计算，带★的要验算．96÷30=252÷80=★235÷76=416÷32=423÷31=★682÷34=”的学库宝（http://www.xuekubao.com/）教师分析与解答如下所示：
【分析】根据整数除法运算的计算法则计算即可求解．注意带★的要验算．
【解答】解：96÷303……2★235÷763…746÷…20★682÷3420…2
【考点】整数的除法及应用．
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知识点讲解
经过分析，习题“列竖式计算，带★的要验算．96÷30=252÷80=★235”主要考察你对
等考点的理解。
因为篇幅有限，只列出部分考点，详细请访问。
整数的除法及应用
1.整数除法：从被除数的最高位除起，除到被除数的哪一位，商就写在那一位上面，每次除后余下的数必须比余数小。2.运算法则： a÷b÷c=a÷（b×c） a÷b×c=a÷（b÷c） a÷（b×c）＝a÷b÷ca÷（b÷c）＝a÷b×c a÷b×c=a×c÷b
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1&&&&2&&&&3&&&&4&&&&5&&&&6&&&&7&&&&8&&&&9&&&&10&&&&11&&&&12&&&&13&&&&14&&&&15&&&&
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3亿+用户的选择
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3亿+用户的选择
计算．验算92+29=验算243+300=验算423+108=验算654-544=验算623-308=验算556-345=（252-188）÷8
708-（271+229）
作业帮用户
扫二维码下载作业帮
3亿+用户的选择
92+29=121验算：243+300=543验算：423+108=531验算：654-544=110验算：623-308=315 验算：556-345=211 验算：（252-188）÷8 =64÷8=8230-6×5 =230-30=200708-（271+229）=708-500=208
为您推荐：
根据整数加减法的计算法则及四则混合运算的运算顺序和计算法则进行计算，注意验算方法的选择．
本题考点：
整数的加法和减法
整数四则混合运算
考点点评：
考查了整数加减法及四则混合运算，注意运算顺序和运算法则，然后再进一步计算．
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Water-and-steam-properties 水和水蒸气性质计算程序 Pro7.0采用国际公认的《工业用1967年IFC公式》和 Other systems 其他 252万源代码下载- www.pudn.com
&文件名称: Water-and-steam-properties& & [
& & & & &&]
&&所属分类:
&&开发工具: Visual Basic
&&文件大小: 1575 KB
&&上传时间:
&&下载次数: 0
&&提 供 者:
&详细说明：水和水蒸气性质计算程序WaterPro7.0采用国际公认的《工业用1967年IFC公式》和《工业用1997年IAPWS公式》进行计算，计算结果在相应版本国际水蒸气热力性质骨架表的允差范围内。
程序能根据压力、温度、焓、熵、比容五个参数(有时加上干度参数)中的任意两个参数，求出相关的15个水蒸汽参数，如定压比热、定容比热、内能、音速、定熵指数、动力粘度、运动粘度、导热系数、普朗特数、介电常数、当给定波长为0.226500 μm时折射率，等等。
-Water and steam properties calculation program WaterPro7.0 adoption of internationally recognized the
industrial 1967 IFC formula
industrial 1997 IAPWS formulas
calculate calculated results in the corresponding version of the steam thermodynamic properties of skeleton table of the status of the poor range.
Program according to the five parameters such as pressure, temperature, enthalpy, entropy, specific volume (sometimes with dry degree parameter) of any two parameters for 15 related to a steam parameters, such as constant pressure specific heat and constant volume specific heat, internal energy, speed, entropy index, dynamic viscosity and kinematic viscosity, heat conduction coefficient, Prandtl number, dielectric electric constant, when the certain wavelength is 0.226500 m refraction rate, and so on.
文件列表(点击判断是否您需要的文件，如果是垃圾请在下面评价投诉):
&&MainFRM.dcu&&MainFRM.ddp&&MainFRM.dfm&&MainFRM.pas&&TS67UE.Tee&&TS67UI.Tee&&TS97UE.Tee&&TS97UI.Tee&&TSIF97.Tee&&TSIFC67.Tee&&WASP.bkm&&WASP.dcu&&WASP.pas&&WASP.res&&WASPS.bkm&&WASPS.cfg&&WASPS.dof&&WASPS.dpr&&WASPS.res&&res&&...\H2O.ico&&...\Thumbs.db&&...\Water.ico&&...\water02.jpg&&About.bkm&&About.dcu&&About.ddp&&About.dfm&&About.pas&&CalChart.bkm&&CalChart.dcu&&CalChart.ddp&&CalChart.dfm&&CalChart.pas&&CalForm.bkm&&CalForm.dcu&&CalForm.ddp&&CalForm.dfm&&CalForm.PAS&&CHelp.dcu&&CHelp.ddp&&CHelp.dfm&&CHelp.pas&&ComMod.bkm&&ComMod.dcu&&ComMod.ddp&&ComMod.dfm&&ComMod.pas&&Config.bkm&&Config.dcu&&Config.ddp&&Config.dfm&&Config.pas&&CRegister.dcu&&CRegister.ddp&&CRegister.dfm&&CRegister.pas&&HS67UE.Tee&&HS67UI.Tee&&HS97UE.Tee&&HS97UI.Tee&&HSIF97.Tee&&HSIFC67.Tee&&IF97.bkm&&IF97.dcu&&IF97.pas&&IFC67.bkm&&IFC67.dcu&&IFC67.pas&&MainFRM.bkm
&输入关键字，在本站252万海量源码库中尽情搜索：PCA82C252和TJA1053容错CAN收发器的计算实例
1. Comparison PCA82C252 / TJA1053 / TJA1054..........................................................................31.1. System parameters ..................................................................................................................31.2. Device parameters ...................................................................................................................32. Series Resistor at Pin BAT of the TJA1053 / TJA1054..............................................................43. Vcc Supply and Recommended Bypass Capacitance ..............................................................53.1. List of used Abbreviations.........................................................................................................53.2. Summary.................................................................................................................................63.3. Average Supply Current at Absence of Bus Short-Circuit Conditions.........................................73.3.1. Maximum dominant supply current (without bus wiring faults) .............................................73.3.2. Example calculation ...........................................................................................................73.3.3. Thermal considerations (without bus wiring faults) ..............................................................73.3.4. Example calculation ...........................................................................................................73.4. Average Supply Current at Presence of a Short-Circuit of one Bus Wire ..................................83.4.1. Maximum dominant supply current (with CANH shorted to GND)........................................83.4.2. Example calculation ...........................................................................................................83.4.3. Thermal considerations (with CANH shorted to GND).........................................................83.4.4. Example calculation ...........................................................................................................83.4.5. Vcc extra supply current in single fault condition.................................................................93.4.6. Example calculation ...........................................................................................................93.5. Worst Case Max Vcc Supply at Presence of a Dual Short Circuit ............................................103.5.1. Max Vcc supply current in worst case dual fault condition.................................................103.5.2. Example calculation .........................................................................................................103.5.3. Vcc extra supply current in dual fault condition .................................................................113.5.4. Example calculation .........................................................................................................113.6. Calculation of worst-case bypass capacitor.............................................................................113.6.1. Example calculation, separate supplied transceiver @ 83,33kBit/s ...................................123.6.2. Example calculation, shared supply..................................................................................124. Calculation of Bus Termination Resistors and EMC issues...................................................134.1. How to dimension the Bus Termination Resistor values, some basic rules ..............................134.1.1. Variable System Size, Optional Nodes .............................................................................134.1.2. Example calculation, Variable System Size ......................................................................144.2. Tolerances of Bus Termination Resistors, EMC Considerations..............................................144.3. Output Current and Power Dissipation of Bus Termination Resistors RT..................................154.3.1. Summary .........................................................................................................................154.3.2. Average power dissipation, no bus failures.......................................................................154.3.3. Example calculation, average power dissipation...............................................................154.3.4. Maximum continuous power dissipation............................................................................154.3.5. Example calculation, maximum continuous power dissipation...........................................154.3.6. Maximum peak power dissipation.....................................................................................164.3.7. Example calculation, maximum peak power dissipation....................................................165. ESD Protection Circuit Concept ..............................................................................................16
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