1=1\'))HAVING9185=3109AND((\'gGeb\'LIKE\'gGeb
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1/11-11/31+2/11-12/31+3/11-13/31+.+10/1-20/31=1/11x(1+10)x10÷2-1/31x(11+20)x10÷2=1/11x11x5-1/31x31x5=5-5=0如果本题有什么不明白可以追问
1/n(n+5)=(1/5)[1/n-1/(n+5)]于是1/6x11=(1/5)x(1/6-1/11)……1/31x36=(1/5)x(1/31-1/36)于是1/6×11+1/11×16+1/16×21+1/21×26+1/26×31+
1-3+2-6+3-9+3-12.+31-93+32-96+33-99=1+2+3.+31+32+33-3-6-9-12-15-18.-99=1+2+3.+31+32+33-3(1+2+3.+31+32+33)=-2(1+2+3.+31+3
一、三、五、七、八、十、十二月每个月有31天
31=thirty-first/31st32=thirty-second/32nd33=thirty-third/33rd34=thirty-fourth/34th35=thirty-fifth/35th
求a的值么?31/3=(16+a)/2-131/3=(16+a-2)/231/3=(14+a)/231X2=3X(14+a)62=42+3a20=3aa=20/3=_=!
4X+3X+3=317X=28X=4
先整理,可以得到2个新式子:3X+28=Y,3X+20=5Y,分别定为3式,4式,再用4式减去3式得4Y=-8,所以Y=-2,代入得到X=-10
1/[n(n+1)]=1/n-1/(n+1)(n=1,2,3.)你自己验证吧
1/2+1/6+1/12.+1/(n-1)n+1/n(n+1)=1/1*2+1/2*3+1/3*4+...+1/(n-1)n+1/n(n+1)=1-1/2+1/2-1/3+1/3-1/4+...+1/(n-1)-1/n+1/n-1/(n+1
10(a+n)-10a+a+n=9n再答:10(a+n)+a-10a+a+n=9n再答:打错再答:你第四个等式打错了再问:哦
1/n(n+2)(n+4)=1/4[1/n(n+2)-1/(n+2)(n+4)]∴原式=1/4*(1/11*13-1/13*15+1/13*15-1/15*17+……+1/29*31-1/31*33)=1/4*(1/11*13-1/31*3
=1/4*(1/11-1/13-1/13+1/15+1/13-1/15-1/15+1/17+...+1/29-1/31-1/31+1/33)中间的全部抵消掉,只剩下首尾的四项1/4*(1/11-1/13-1/31+1/33)最后得到440/
=31/2+21/4-1/3-1/4-1/6=186/12+63/12-4/12-3/12-2/12=240/12=20
这个问题很简单啊,你把所有分母是11的加在一起,把所有分母是31的相加,两者相减,得到的答案是5-5=0.
an=1/[(2^n)-1]
51=1x2x3x4x5=12061=1x2x3x4x5x6=720所以51/61=6分之1
31*39=120935*35=1225分别除以70整数部分为17给力采纳
等于(1-1/2)+(1-1/4)+(1-1/8)+(1-1/16)+(1-1/32)+(1-1/64)+(1-1/128)=7-(1/2+1/4+1/8+1/16+1/32+1/64+1/128)=7-127/128=6又1/128