3+4=12014年8月1日是星期几,4+8=1年,8+5=?

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空气污染扩散指数
气象条件有利于空气污染物扩散。
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涂擦SPF大于15、PA+防晒护肤品。
感冒机率较低,避免长期处于空调屋中。
有雨,雨水和泥水会弄脏爱车。
有降水,推荐您在室内进行休闲运动。
空气污染扩散指数
气象条件有利于空气污染物扩散。
紫外线指数
涂擦SPF大于15、PA+防晒护肤品。
感冒机率较低,避免长期处于空调屋中。
有雨,雨水和泥水会弄脏爱车。
有降水,推荐您在室内进行休闲运动。
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有雨,雨水和泥水会弄脏爱车。
有降水,推荐您在室内进行休闲运动。
空气污染扩散指数
气象条件有利于空气污染物扩散。
紫外线指数
涂擦SPF大于15、PA+防晒护肤品。
感冒机率较低,避免长期处于空调屋中。
无雨且风力较小,易保持清洁度。
推荐您在室内进行低强度运动。
空气污染扩散指数
气象条件有利于空气污染物扩散。
紫外线指数
辐射弱,涂擦SPF8-12防晒护肤品。
感冒机率较低,避免长期处于空调屋中。
有降水,推荐您在室内进行休闲运动。
空气污染扩散指数
气象条件有利于空气污染物扩散。
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Copyright& All Rights Reserved () 版权所有 复制必究 郑重声明:中国天气网版权所有,未经书面授权禁止使用已知全集U={1,2,3,4,5,6,7,8},A={x|x2-3x+2=0},B={x|1≤x≤5
提问:级别:二年级来自:安徽省合肥市
回答数:2浏览数:
已知全集U={1,2,3,4,5,6,7,8},A={x|x2-3x+2=0},B={x|1≤x≤5
已知全集U={1,2,3,4,5,6,7,8},A={x|x2-3x+2=0},B={x|1≤x≤5,x属于Z}C={x|2&x&9,x属于Z}
求A∪(B∩C)
求(CUB)∪(CUC).
请尽量给出过程
&提问时间: 16:53:33
最佳答案此答案已被选择为最佳答案,但并不代表问吧支持或赞同其观点
回答:级别:高二 12:32:34来自:河南省平顶山市
先解方程得到:A={1,2},根据后面范围得B={1,2,3,4,5},C={3,4,5,6,7,8}
然后就知道A∪(B∩C)=={1,2}∪{3,4,5}={1,2,3,4,5},
第二个式子可以先变形,利用性质得到={1,2,6,7,8}
提问者对答案的评价:
回答:级别:高级教员 17:32:21来自:山东省临沂市
由题意可知A={1,2}
B={1,2,3,4,5}
C={3,4,5,6,7,8}
所以B∩C={3,4,5}
所以A∪(B∩C)={1,2,3,4,5}
CUB={6,7,8}
所以(CUB)∪(CUC)={1,2,6,7,8}
总回答数2,每页15条,当前第1页,共1页
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最新热点问题分析:(1)利用二次函数的性质,结合函数图象可求(2)要求原函数的值域,转化为求二次函数-x2-6x-5的值域问题的求解,基本方法是配方((3)把函数化简y=3x+1x-2=3(x-2)+7x-2=3+7x-2,结合反比例函数的性质可求(4)利用换元法,然后结合二次函数的性质可求函数的值域.(5)利用换元,令x=cosα,然后由辅助角公式,结合正弦函数的性质可求(6)利用分段函数进行讨论,把函数化简为y=|x-1|+|x+4|=2x+3,x≥15,-4<x<1-2x-3,x≤-4,从而可求(7)利用判别式法进行求解(8)由y=(x-12)2+12(x-12)+12x-12,分离系数后利用基本不等式求解函数的值域(9)由于y=1-sinx2-cosx=sinx-1cosx-2可以看着在单位圆上任取一点与定点A(2,1)的连线的斜率,根据几何意义可求函数的值域(10)利用分离系数法,结合反比例函数的值域进行求解(11)利用换元,结合二次函数的配方法进行求解(12)分x>0,x=0,x<0三种情况,分子分母同时x,然后结合二次函数的配方法进行求解(13)利用二次函数的配方法进行求解函数的值域(14)利用函数的单调性进行求解函数的值域(15)利用分离系数法,然后由二次函数的值域的求解的配方法进行求解解答:解(1)y=3x2-x+2由二次函数的性质可知,当x=16时,函数有最小值2312故函数的值域为[2312,+∞)(2)y=-x2-6x-5=-(x+3)2+4∵0≤-(x+3)2+4≤40∴0≤y≤2故函数的值域[0,2](3)y=3x+1x-2=3(x-2)+7x-2=3+7x-2≠3故函数的值域(-∞,3)∪(3,+∞)(4)令1-x=t则t≥0且x=1-t2y=x+41-x=1-t2+4t=-(t-2)2+5在[0,2]上单调递增,在[2,+∞)单调递减当t=2时,函数有最大值5∴函数的值域为(-∞,5](5)令x=cosα,则y=x+1-x2=cosα+sinα=2sin(α+π4)∴-2≤y≤&2(6)y=|x-1|+|x+4|=2x+3,x≥15,-4<x<1-2x-3,x≤-4∴y≥5故函数的值域[5,+∞)(7)∵y=2x2-x+2x2+x+1∴(y-2)x2+(y+1)x+y-2=0①当y=2时,x=0满足条件②当y≠2时,△=(y+1)2-4(y-2)2≥0即y2-6y+5≤0解可得1≤y≤5且y≠2综上可得,1≤y≤5故函数的值域为{y|1≤y≤5}&(8)∵x>12∴x-12>0∴x-12+12x-12≥2(x-12)&#=2∴y=(x-12)2+12(x-12)+12x-12=x+12+12x-12+12≥2+12故函数的值域为[2+12,+∞)(9)∵y=1-sinx2-cosx=sinx-1cosx-2可以看着在单位圆上任取一点与定点A(2,1)的连线的斜率当直线与圆相切时,由圆心到直线的距离为半径可得斜率k=0或k=43∴0≤k≤43故函数的值域为[0,43](10)∵y=x2-5x+6x2+x-6=(x-2)(x-3)(x+3)(x-2)=x-3x+3(x≠2)∴y=x-3x+3=x+3-6x+31-6x+3∴y≠-15且y≠1∴函数的值域为{y|y≠1且y≠-15}(11)∵y=2x+41-x令1-x=t,则x=1-t2且t≥0∴y=2x+41-x=2(1-t2)+4t=-2t2+4t+2=-2(t-1)2+4根据二次函数的 性质可知,当t=1时,函数有最大值4函数的值域为(-∞,4](12)y=-xx2+2x+2①当x=0时,y=0②当x>0,y=-xx2+2x+2=-1x2+2x+2x2=-11+2x+2x2∵2x2+2x+1=2(1x+12)2+12>1∴y>-1③当x<0时,y=-xx2+2x+2=11+2x+2x2∵2x2+2x+1=2(1x+12)2+12≥12∴y≤2综上可得,函数的值域为R(13)∵y=4-3+2x-x2的定义域[-1,3]令f(x)=-x2+2x+3=-(x-1)2+4则0≤f(x)≤4∴0≤3+2x-x2≤2∴2≤f(x)≤4即函数的值域[2,4](14)∵y=x-1-2x的定义域为(-∞,12],且在(-∞,12]上单调递增∴当x=12时,函数有最大值12故函数的值域(-∞,12](15)∵y=2x2+2x+5x2+x+1∴(y-2)x2-(y-2)x+y-5=0∴△=(y-2)2-4(y-2)(y-5)≥0即(y-2)(3y-18)≤0∴2≤y≤6故函数的值域(2,6]点评:本题主要考查了函数值域求解的一些常用方法的应用,要注意配方、换元、函数的单调性、判别式法、及利用几何意义等方法的应用
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Can anyone help please? I need to know how to make the sum of value S(n) = 0+1+2+3+4...+(n-1)+n (example: 0+1+2+3 = 6 , 0+1+2+3+4+5+6+7 = 28 and so on) on Latex
110k9193389
The difficult task is generating the terms of the sequence, not computing the sum, I present a macro that prints all the terms or just the sum. You can define a different starting point and another difference (defaults 0 and 1).
\documentclass{article}
\usepackage{xparse}
\ExplSyntaxOn
\NewDocumentCommand{\arithmeticsequence}{sO{}m}
\group_begin:
\keys_set:nn {rudstep/arseq} { #2 }
\IfBooleanTF{#1}
{ \rudstep_arseq_sum:n { #3 } }
{ \rudstep_arseq_full:n { #3 } }
\group_end:
\keys_define:nn { rudstep/arseq }
.int_set:N = \l_rudstep_diff_int,
start .int_set:N = \l_rudstep_start_int,
.initial:n = 1,
start .initial:n = 0,
\seq_new:N \l_rudstep_terms_seq
\cs_new_protected:Npn \rudstep_arseq_full:n #1
\seq_clear:N \l_rudstep_terms_seq
\int_step_inline:nnnn
{ \l_rudstep_start_int } % start
{ \l_rudstep_diff_int }
{ \l_rudstep_start_int + #1*\l_rudstep_diff_int }
{ \seq_put_right:Nn \l_rudstep_terms_seq { ##1 } }
$\seq_use:Nn \l_rudstep_terms_seq { + } = \rudstep_arseq_sum:n { #1 }$
\cs_new:Npn \rudstep_arseq_sum:n #1
\int_eval:n { (#1+1)*(2*\l_rudstep_start_int+#1*\l_rudstep_diff_int)/2 }
\ExplSyntaxOff
\begin{document}
Just the sum of five terms after 0: \arithmeticsequence*{5}
The whole sequence: \arithmeticsequence{3}
Or: \arithmeticsequence{4}
Start from 1: \arithmeticsequence[start=1]{3}
A big example:
\arithmeticsequence[start=81297,diff=198]{180}
\end{document}
Of course, if you know how to do it in Lisp, then here's the way:
\documentclass{article}
\usepackage{lisp-on-tex}
\begin{document}
\lispinterp{
(\define \intsum
(\lambda (\n)
(\lispif (\= \n :0 ) :0
(\+ (\intsum (\- \n :1)) \n))))}
\lispinterp{(\texprint(\intsum:100))}
\end{document}
that prints, as is well known,
Note: shamelessly adapted from the documentation of lisp-on-tex.
The e-TeX version of the same idea, using recursion, can be
\documentclass{article}
\newcommand{\Sn}[1]{%
\number\numexpr #1
\ifnum\numexpr#1-1&0
+\expandafter\Sn\expandafter{\number\numexpr#1-1}
\begin{document}
\end{document}
Note that this is fully expandable. The macro \Sn starts from the argument and if it's bigger than 1 asks to expand \Sn with the argument decreased by 1. One might start also from 0, but I'll leave it as an exercise to the interested reader.
494k5613312346
\documentclass{article}
\newcommand\foo[1]{\the\numexpr((#1)*(#1+1))/2\relax}
% or if you want to print the terms
\newcommand\foob[1]{$\fooc{#1}{0}=\foo{#1}$}
\newcommand\fooc[2]{\the\numexpr#2\relax\ifnum#1=#2\relax\else+\fooc{#1}{\numexpr#2+1\relax}\fi}
\begin{document}
\foo{8} and \foo{5}
\foob{8} and \foob{5}
\end{document}
333k267971407
Using Lua is probably a huge overkill in this situation, but it shows off how one can easily integrate Lua in LaTeX.
The code might also be easier to grasp for programmers who are beginners in LaTeX ;)
% !TEX TS-program = lualatex
% !TEX encoding = UTF-8 Unicode
\documentclass{article}
\usepackage[utf8]{luainputenc}
\usepackage{luacode}
% The code won't work as-is if you move it into \directlua{}.
% If you do so, be sure to replace "~" (in i ~= startInt) by \string~
\begin{luacode}
function calcSum(startInt, endInt)
local sum = 0
for i=startInt, endInt do
sum = sum + i
return sum
function writeSum(startInt, endInt)
local sum = 0
local tex = ""
for i=startInt, endInt do
if i ~= startInt then
tex = tex .. " + "
sum = sum + i
tex = tex .. i
tex = tex .. " = " .. sum
return tex
\end{luacode}
\def\calcSum#1#2{\directlua{
tex.print(calcSum(tonumber(#1), tonumber(#2)))
\def\writeSum#1#2{\directlua{
tex.print(writeSum(tonumber(#1), tonumber(#2)))
\begin{document}
The sum of the numbers \(0, 1, 2, \ldots, 28\) is \(\calcSum{0}{28}\).\\
Another sum: \(\writeSum{100}{103}\).
\end{document}
Here is a short expl3 approach using \int_step_inline:nnnn. This carries out the explicit sum and doesn't rely on Gauss' reduction. So this is for illustration purposes only.
\documentclass{article}
\usepackage{xparse}
\ExplSyntaxOn
\cs_generate_variant:Nn \int_to_arabic:n { V }
\cs_new:Npn \rudstep_sum:n #1
\int_zero:N \l_tmpa_int
\int_step_inline:nnnn { 1 } { 1 } { #1 }
\int_add:Nn \l_tmpa_int { ##1 }
\int_to_arabic:V \l_tmpa_int
\NewDocumentCommand \calcsum {m}
\rudstep_sum:n { #1 }
\ExplSyntaxOff
\begin{document}
\calcsum{8}
\end{document}
27.2k365146
Using my package calculator, you can perform arithmetic calculations comfortably. This code solves your problem:
\documentclass{article}
\usepackage{ifthen}
\usepackage{calculator}
\newcounter{n}
\newcommand{\sumZeroToN}[2]{%
\COPY{0}{#2}
\whiledo{\not{\value{n}&#1}}{%
\ADD{#2}{\value{n}}{#2}\stepcounter{n}}}
\begin{document}
\sumZeroToN{100}{\sol}
\sum_{n=0}^{100} n = \sol
\end{document}
An expandable solution using the package
for computations and \xintiloop for expandably looping with an index.
\documentclass{article}
\usepackage{bnumexpr}% minimally extends \numexpr to big integers (new on CTAN
% for \xintiloop: (for fun)
\usepackage{xinttools}% automatically loaded by xint, which is loaded by
% bnumexpr, but in the future bnumexpr will only load a
% smaller part of xint, and xint itself will not load
% xinttools anymore.
\newcommand{\ArithmeticSequenceAndItsSum}[3]{%
% #1= initial term, A
% #2= common difference, d
% #3= number of terms, N
% sum is A+(A+d)+...+(A+(N-1)d)=N*A+d*N(N-1)/2
% (cf. Gauss Werke, Kindergarten Abteilung)
#1% assume N at least one and that #1 does not need to be evaluated
\xintiloop [1+1]
\unless\ifnum#3=\xintiloopindex
+\thebnumexpr#1+\xintiloopindex*#2\relax
=\thebnumexpr #3*#1+#2*#3*(#3-1)/2\relax
\begin{document}
$\ArithmeticSequenceAndItsSum {0}{1}{10}$
$\ArithmeticSequenceAndItsSum {1}{2}{10}$
\noindent $\ArithmeticSequenceAndItsSum {}{}{100}$
And now go see the log for some bigger integers!!
\message{\ArithmeticSequenceAndItsSum {}{}{1000}}
\end{document}
29.4k46101
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TeX - LaTeX Stack Exchange works best with JavaScript enabled4.7+2.3=4.5×2=6.9-2.5=7.2×0.8=6×3.4=0.62-0.32=1.4×0.5=0.75×100=0.02×0.5=3.6÷0.3=6.3÷7=5.6÷100=0.75÷0.25=0.125×8=4.8÷0.3=0.96÷2=0.56÷28=0.36÷0.4=0.64÷0.8=0.72÷3.6=3.6÷24=0.8×1.1=7.2+12.8=46.7-3.8=12.8÷4=5.2÷13=12.5÷5=1.64÷41=10÷20=24÷15=8.65×10=0.35×0.6=3.08×0.01=4.95××0.1=0.4×0.5=2.4÷0.8=10.8÷9=9.6÷0.8=0.108÷2=4.95÷0.9=9.65÷0.1=0.325×100=2.5×8=0.56÷0.7=0.125×4=3.28×0.1=3.9÷0.13=7.2×0.1=0.01×0.1=0.25×0.4=1.6÷0.8=1÷2.5=1.25×0.8=3.2÷0.04=0÷1.7=0.22×102=9.6÷0.8=5×0.24=4.5-0.05=3.9÷0..5=5×0.12=24×0.5=4.8×0.5=2.8+4.2=0.84÷2.1=5÷0.25=7.8÷0.01=5.4÷0.6=3.2÷5=7×0.62=0.56÷0.8=7.4-2.8=0.18÷0.2=0.16÷8=4.5×0.02=8+7.2=1.2×30=0.012×0.2=7.3×0.3=9÷6=3.6÷0.12=7.8÷0.6=2÷5=4.5+5=0.3÷0.6=0.96÷0.2=10.5×0.4=7.3+0.27=8×0.125=0.54÷0.6=0.61+0.39=0.56÷28=10÷20=4.08÷0.4=11-9.4=0.8×0.11=0.72+12.8=5.6÷0.01=2.3×100=0.75×100=0.108÷2=10.8÷9=0.7÷0.8=16-5.07=4.2×0.4=0.72×6=12÷0.5=5.2×0.4=12.2÷0.2=5.6÷100=0.41+3.7=0.02×0.5=7.2-0.8=1.4×0.5=84÷0.21=0.75÷0.25=2.5×16=0.108÷2=1.75+32.5=16-5.07=5.2÷13=1.64÷41=1.02×0.2=0.26×0.3=8.4×0.02=1.29×6=0.8×0.05=12.6÷0.03=8.71÷0.1=21÷0.21=0.8×0.13=0.34÷17=8.08÷0.4=0.5×2.2=1.24÷4=3.27+0.63=5.02×0.3=0.3×0.5=6.4÷4=7.2÷1.8=0.7÷0.35=0.72÷1.44=0.48÷0.04=1.25÷2.5=5.5+55=16.8÷8=0.54+2.2=3.5-0.05=2÷0.02=0.25×40=1.5×0.6=0.32÷0.8=12.25÷0.5=73.5×0.1=46.5+52.5=0.45×102=1.25×88=2.64+3.85+1.54=8×1.25=5.6÷3.5=4.2÷0.7÷6=0.4×8.6×25=0.27÷0.3=2.5×101=0.65×101-0.65=2.6×7÷2.6×7= - 跟谁学
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在线咨询您好,告诉我您想学什么,15分钟为您匹配优质老师哦马上咨询&&&分类:4.7+2.3=4.5×2=6.9-2.5=7.2×0.8=6×3.4=0.62-0.32=1.4×0.5=0.75×100=0.02×0.5=3.6÷0.3=6.3÷7=5.6÷100=0.75÷0.25=0.125×8=4.8÷0.3=0.96÷2=0.56÷28=0.36÷0.4=0.64÷0.8=0.72÷3.6=3.6÷24=0.8×1.1=7.2+12.8=46.7-3.8=12.8÷4=5.2÷13=12.5÷5=1.64÷41=10÷20=24÷15=8.65×10=0.35×0.6=3.08×0.01=4.95××0.1=0.4×0.5=2.4÷0.8=10.8÷9=9.6÷0.8=0.108÷2=4.95÷0.9=9.65÷0.1=0.325×100=2.5×8=0.56÷0.7=0.125×4=3.28×0.1=3.9÷0.13=7.2×0.1=0.01×0.1=0.25×0.4=1.6÷0.8=1÷2.5=1.25×0.8=3.2÷0.04=0÷1.7=0.22×102=9.6÷0.8=5×0.24=4.5-0.05=3.9÷0..5=5×0.12=24×0.5=4.8×0.5=2.8+4.2=0.84÷2.1=5÷0.25=7.8÷0.01=5.4÷0.6=3.2÷5=7×0.62=0.56÷0.8=7.4-2.8=0.18÷0.2=0.16÷8=4.5×0.02=8+7.2=1.2×30=0.012×0.2=7.3×0.3=9÷6=3.6÷0.12=7.8÷0.6=2÷5=4.5+5=0.3÷0.6=0.96÷0.2=10.5×0.4=7.3+0.27=8×0.125=0.54÷0.6=0.61+0.39=0.56÷28=10÷20=4.08÷0.4=11-9.4=0.8×0.11=0.72+12.8=5.6÷0.01=2.3×100=0.75×100=0.108÷2=10.8÷9=0.7÷0.8=16-5.07=4.2×0.4=0.72×6=12÷0.5=5.2×0.4=12.2÷0.2=5.6÷100=0.41+3.7=0.02×0.5=7.2-0.8=1.4×0.5=84÷0.21=0.75÷0.25=2.5×16=0.108÷2=1.75+32.5=16-5.07=5.2÷13=1.64÷41=1.02×0.2=0.26×0.3=8.4×0.02=1.29×6=0.8×0.05=12.6÷0.03=8.71÷0.1=21÷0.21=0.8×0.13=0.34÷17=8.08÷0.4=0.5×2.2=1.24÷4=3.27+0.63=5.02×0.3=0.3×0.5=6.4÷4=7.2÷1.8=0.7÷0.35=0.72÷1.44=0.48÷0.04=1.25÷2.5=5.5+55=16.8÷8=0.54+2.2=3.5-0.05=2÷0.02=0.25×40=1.5×0.6=0.32÷0.8=12.25÷0.5=73.5×0.1=46.5+52.5=0.45×102=1.25×88=2.64+3.85+1.54=8×1.25=5.6÷3.5=4.2÷0.7÷6=0.4×8.6×25=0.27÷0.3=2.5×101=0.65×101-0.65=2.6×7÷2.6×7=4.7+2.3=4.5×2=6.9-2.5=7.2×0.8=6×3.4=0.62-0.32=1.4×0.5=0.75×100=0.02×0.5=3.6÷0.3=6.3÷7=5.6÷100=0.75÷0.25=0.125×8=4.8÷0.3=0.96÷2=0.56÷28=0.36÷0.4=0.64÷0.8=0.72÷3.6=3.6÷24=0.8×1.1=7.2+12.8=46.7-3.8=12.8÷4=5.2÷13=12.5÷5=1.64÷41=10÷20=24÷15=8.65×10=0.35×0.6=3.08×0.01=4.95×1000=6.9×0.1=0.4×0.5=2.4÷0.8=10.8÷9=9.6÷0.8=0.108÷2=4.95÷0.9=9.65÷0.1=0.325×100=2.5×8=0.56÷0.7=0.125×4=3.28×0.1=3.9÷0.13=7.2×0.1=0.01×0.1=0.25×0.4=1.6÷0.8=1÷2.5=1.25×0.8=3.2÷0.04=0÷1.7=0.22×102=9.6÷0.8=5×0.24=4.5-0.05=3.9÷0.1316.5÷0.5=5×0.12=24×0.5=4.8×0.5=2.8+4.2=0.84÷2.1=5÷0.25=7.8÷0.01=5.4÷0.6=3.2÷5=7×0.62=0.56÷0.8=7.4-2.8=0.18÷0.2=0.16÷8=4.5×0.02=8+7.2=1.2×30=0.012×0.2=7.3×0.3=9÷6=3.6÷0.12=7.8÷0.6=2÷5=4.5+5=0.3÷0.6=0.96÷0.2=10.5×0.4=7.3+0.27=8×0.125=0.54÷0.6=0.61+0.39=0.56÷28=10÷20=4.08÷0.4=11-9.4=0.8×0.11=0.72+12.8=5.6÷0.01=2.3×100=0.75×100=0.108÷2=10.8÷9=0.7÷0.8=16-5.07=4.2×0.4=0.72×6=12÷0.5=5.2×0.4=12.2÷0.2=5.6÷100=0.41+3.7=0.02×0.5=7.2-0.8=1.4×0.5=84÷0.21=0.75÷0.25=2.5×16=0.108÷2=1.75+32.5=16-5.07=5.2÷13=1.64÷41=1.02×0.2=0.26×0.3=8.4×0.02=1.29×6=0.8×0.05=12.6÷0.03=8.71÷0.1=21÷0.21=0.8×0.13=0.34÷17=8.08÷0.4=0.5×2.2=1.24÷4=3.27+0.63=5.02×0.3=0.3×0.5=6.4÷4=7.2÷1.8=0.7÷0.35=0.72÷1.44=0.48÷0.04=1.25÷2.5=5.5+55=16.8÷8=0.54+2.2=3.5-0.05=2÷0.02=0.25×40=1.5×0.6=0.32÷0.8=12.25÷0.5=73.5×0.1=46.5+52.5=0.45×102=1.25×88=2.64+3.85+1.54=8×1.25=5.6÷3.5=4.2÷0.7÷6=0.4×8.6×25=0.27÷0.3=2.5×101=0.65×101-0.65=2.6×7÷2.6×7=科目:最佳答案4.7+2.3=74.5×2=96.9-2.5=4.47.2×0.8=5.766×3.4=20.40.62-0.32=0.31.4×0.5=0.70.75×100=750.02×0.5=0.013.6÷0.3=126.3÷7=0.95.6÷100=0.0560.75÷0.25=30.125×8=14.8÷0.3=160.96÷2=0.480.56÷28=0.020.36÷0.4=0.90.64÷0.8=0.80.72÷3.6=0.23.6÷24=0.150.8×1.1=0.887.2+12.8=2046.7-3.8=42.912.8÷4=3.25.2÷13=0.412.5÷5=2.51.64÷41=0.0410÷20=0.524÷15=1.68.65×10=86.50.35×0.6=0.213.08×0.01=0.03084.95×6.9×0.1=0.690.4×0.5=0.22.4÷0.8=310.8÷9=1.29.6÷0.8=120.108÷2=0.0544.95÷0.9=5.59.65÷0.1=96.50.325×100=32.52.5×8=200.56÷0.7=0.80.125×4=0.53.28×0.1=0.3283.9÷0.13=307.2×0.1=0.720.01×0.1=0.0010.25×0.4=0.11.6÷0.8=21÷2.5=0.41.25×0.8=13.2÷0.04=800÷1.7=00.22×102=22.449.6÷0.8=125×0.24=1.24.5-0.05=4.453.9÷0.13=3016.5÷0.5=335×0.12=0.624×0.5=124.8×0.5=2.42.8+4.2=70.84÷2.1=0.45÷0.25=207.8÷0.01=7805.4÷0.6=93.2÷5=0.647×0.62=4.340.56÷0.8=0.77.4-2.8=4.60.18÷0.2=0.90.16÷8=0.024.5×0.02=0.098+7.2=15.21.2×30=360.012×0.2=0.00247.3×0.3=2.199÷6=1.53.6÷0.12=307.8÷0.6=132÷5=0.44.5+5=9.50.3÷0.6=0.50.96÷0.2=4.810.5×0.4=4.27.3+0.27=7.578×0.125=10.54÷0.6=0.90.61+0.39=10.56÷28=0.0210÷20=0.54.08÷0.4=10.211-9.4=1.60.8×0.11=0.0880.72+12.8=13.525.6÷0.01=5602.3×100=2300.75×100=750.108÷2=0.05410.8÷9=1.20.7÷0.8=0.87516-5.07=10.934.2×0.4=1.680.72×6=4.3212÷0.5=245.2×0.4=2.0812.2÷0.2=615.6÷100=0.0560.41+3.7=4.110.02×0.5=0.017.2-0.8=6.41.4×0.5=0.784÷0.21=4000.75÷0.25=32.5×16=400.108÷2=0.0541.75+32.5=34.2516-5.07=10.935.2÷13=0.41.64÷41=0.041.02×0.2=0.2040.26×0.3=0.0788.4×0.02=0.1681.29×6=7.740.8×0.05=0.0412.6÷0.03=4208.71÷0.1=87.121÷0.21=1000.8×0.13=0.1040.34÷17=0.028.08÷0.4=20.20.5×2.2=1.11.24÷4=0.313.27+0.63=3.95.02×0.3=1.5060.3×0.5=0.156.4÷4=1.67.2÷1.8=40.7÷0.35=20.72÷1.44=0.50.48÷0.04=121.25÷2.5=0.55.5+55=60.516.8÷8=2.10.54+2.2=2.743.5-0.05=3.452÷0.02=1000.25×40=101.5×0.6=0.90.32÷0.8=0.412.25÷0.5=24.573.5×0.1=7.3546.5+52.5=990.45×102=45.91.25×88=1102.64+3.85+1.54=8.038×1.25=105.6÷3.5=1.64.2÷0.7÷6=10.4×8.6×25=8.60.27÷0.3=0.92.5×101=252.50.65×101-0.65=652.6×7÷2.6×7=49解析
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