简算(1+1/2+1/3+1/4)*(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)*(1/2+1/3+1/4)
来源:百度知道 编辑:UC知道 时间:2024/05/24 00:56:07
令a=1+1/2+1/3+1/4
则1/2+1/3+1/4+1/5=a-1+1/5
1+1/2+1/3+1/4+1/5=a+1/5
1/2+1/3+1/4=a-1
所以原式=a(a-1+1/5)-(a+1/5)(a-1)
=a^2-(1-1/5)a-a^2+(1-1/5)a+1/5
=1/5
设1/2+1/3+1/4=x,1+1/2+1/3+1/4+1/5=y
原式=(x+1)(y-1)-xy
=xy+y-x-1-xy
=y-x-1
∵y-x=1+1/5
∴原式=1+1/5-1=1/5
原式=(1+1/2+1/3+1/4)*(1/2+1/3+1/4+1/5)-(1+1/2+1/3+1/4+1/5)*(1/2+1/3+1/4)=(1+1/2+1/3+1/4+1/5)(1+1/2+1/3+1/4-1/2-1/3-1/4)=1+1/2+1/3+1/4+1/5
简算1+2+3+4+5......+?=66
递等式(简算):1+2+3.....+40
(1/2005-1)(1/2004-1)........(1/3-1)(1/2-1)
1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+100)
1+1/(1+2)+1/(1+2+3)+-------+1/(1+2+3+----+100)
1+1/1+2+1/1+2+3+...+1/1+2+3...+2000
1+1/1+2+1/1+2+3.........+1/1+2+3.....100
1*(1/1+2)*(1/1+2+3)*~~~*(1/1+2+~~~2005)=?
(1-1/2)(1-1/3)(1-1/4)(1-1/5).....(1-1/1000)
简算:求1÷(1/1989+1/1990+…+1/2004+1/2005)的整数部分