在求1-1/3+1/5-1/7+一直到某一项绝对值的值小于1e-6（求它们的和）我是这样求得，但为什么结果是0.000000_百度知道
在求1-1/3+1/5-1/7+一直到某一项绝对值的值小于1e-6（求它们的和）我是这样求得，但为什么结果是0.000000
#include &stdio.h&#include &math.h&void main(){ double sum=0; int i=1; while(1/(2*i-1)&=1e-6)
sum=sum+pow(-1,i+1)*1.0/(2*i-1); printf(&%f&,4*sum);}
提问者采纳
#include &stdio.h&#include &math.h&void main(){ double sum=0; int i=1; while(1.0/(2*i-1)&=1e-6)
//1改为1.0 ，&=改为&= {
sum=sum+pow(-1,i+1)*1.0/(2*i-1);
//加一个i++ } printf(&%f\n&,4*sum);}
提问者评价
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其他4条回答
你应该while(1.0/(2*i-1)&=1e-6)，整数相除损失了精度，转换成浮点才行
如果那样的话，系统会提示有错误的！
while(1.0/(2*i-1)&=1e-6){
sum +=pow(-1,i+1)*1.0/(2*i-1);
(2*i-1)表示有问题：pow(2,i)i得自++
我后来又加了i++，但为什么结果又是0.0000000
while(1/(2*i-1)&=1e-6)
{ sum=sum+pow(-1,i+1)*1.0/(2*i-1);
i++;}printf(&%f&,sum);这样试试
当然是错的啊，你循环的条件都写错了啊，你把小于号改成大于号就好了
楼上 春雨_润物 已经概括的很好了，但有一点需要进行修改，楼主最好使用标志变量，而不是使用数学函数来实现正负号的改变，另外可以发现楼主对于循环语句运用的还不太熟练，while，do-while ,以及for都是在满足条件时才进行循环。下面是我写的一个程序，有不足之处望指出： #include &stdio.h&#include &math.h&void main(){ double PI=0; int i=1,leap=1; while(1.0/(2*i-1)&=1e-6)
//1改为1.0 目的是自动类型转换，&=改为&= {
PI+=leap*(1.0/(2*i-1));
//加一个i++ printf(&%.6f\n&,4*PI);}
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出门在外也不愁已知三个不等式|x-3|+1&x,(6x+7/x+2)-2≥x,2x²-3mx+m²&0的解集分别为A,B,C_百度知道
已知三个不等式|x-3|+1&x,(6x+7/x+2)-2≥x,2x²-3mx+m²&0的解集分别为A,B,C
(1)求集合A（2）求集合B（3）若A∩B包含于C求实数m取值范围
提问者采纳
等式1：若x&=3则原式 x-2&x,恒立若x&3则原式4-x&x,x&2,所2&x&3
则集合A{x|x&2}等式2：化x^2-2x-3&=0,所集合B{x|-1&=x&=3}等式3：△=（3m）^2-8m^2=m^2&0
（2x-m）(x-m)&0
若m&0,则集合C{x|m/2&x&m}
若m&0,则集合C{x|m&x&m/2}
m=0解A∩B={x|2&x&=3}
若A∩B包含于Cm&0
m/2&2且m&3
所3&m&4m&0
m&2且m/2&3,解所m取值范围3&m&4
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其他1条回答
等式1：若x&=3则原式 x-2&x,恒立若x&3则原式4-x&x,x&2,所2&x&3
则集合A{x|x&2}等式2：化x^2-2x-3&=0,所集合B{x|-1&=x&=3}等式3：△=（3m）^2-8m^2=m^2&0
（2x-m）(x-m)&0
若m&0,则集合C{x|m/2&x&m}
若m&0,则集合C{x|m&x&m/2}
m=0解A∩B={x|2&x&=3}
若A∩B包含于Cm&0
m/2&2且m&3
所3&m&4m&0
m&2且m/2&3,解所m取值范围3&m&4 望采纳
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出门在外也不愁setuptools 11.3.1
Easily download, build, install, upgrade, and uninstall Python packages
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The recommended way to bootstrap setuptools on any system is to download
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You must start the Powershell with Administrative privileges or you may choose
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& (Invoke-WebRequest https://bootstrap.pypa.io/ez_setup.py).Content | python - --user
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Copyright (C) ,化简（0.064）^-1/3-(-7/8)^0+[(-2)^3]^-4/3+16^-0.75+|-0.01|^1/2,拜求大神解答！！！_百度知道
化简（0.064）^-1/3-(-7/8)^0+[(-2)^3]^-4/3+16^-0.75+|-0.01|^1/2,拜求大神解答！！！
化简（0.064）^-1/3-(-7/8)^0+[(-2)^3]^-4/3+16^-0.75+|-0.01|^1/2
提问者采纳
注意楼上的错了解：原式=（0.4³）^-1/3-1+(-2)^-4+2(⁴x-0.75)+0.01^1/2=0.4-¹ -1+1/16+2-³+0.1=5/2-1+1/16+1/8+0.1=3/2+3/16+1/10=27/16+1/10=135/80+8/80=143/80
额，不是计算，能帮忙化简下吗？（0.064）^-1/3-(-7/8)^0+[(-2)^3]^-4/3+16^-0.75+|-0.01|^1/2
上面就是化简过程本题没有字母，所以化简也就是计算
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感激不尽！！！
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原式=（0.4）^（-1）-1+（-2）^（-4）+（1/16）^（3/4）+√（0.01）
=2.5-1+（1/16）+（1/2）^3+0.1
=1.5-1+（1/16）+（1/8）+0.1
=0.6+（3/16）
√2/2^1/2=?
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