漫画-明太子 第一:求和:1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+1/[(x+3)(x+4)]+1/[(x+4)(x+5)]
来源:百度文库 编辑:高校问答 时间:2024/10/06 17:30:53
希望大家能帮我一下!谢谢了!
1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+1/[(x+3)(x+4)]+1/[(x+4)(x+5)]
=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+.....1/(x+3)-1/(x+4)+1/(x+4)-1/(x+5)
=1/(x+1)-1/(x+5)
=4/[(x+1)(x+5)]
这就是拆项法
这么多人都会啊
那我就不答了
1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+1/[(x+3)(x+4)]+1/[(x+4)(x+5)]
=[1/(x+1)-1/(x+2)]+[1/(x+2)-1/(x+3)]+[1/(x+3)-1/(x+4)]+[1/(x+4)-1/(x+5)]
=1/(x+1)-1/(x+5)
=4/[(x+1)(x+5)]
最烦这种没有用的数学,中国的教育体制啊!无言!...
曾经会做!现在不会了阿!!初中的!这还是最基本的代数相乘!
这是哪个年级的数学?
求和:1/[(x+1)(x+2)]+1/[(x+2)(x+3)]+1/[(x+3)(x+4)]+1/[(x+4)(x+5)]
求和:1/x(x+3)+1/(x+3)(x+6)+......+1/(x+2001)(x+2004)
求和S=1+2x+3x平方+...+nx的(n-1)次方
(x*x*x-x*x+x)/(x*x*x+1)+(x*x+x+1)/[(1-x*x)/x)]
(x*x*x-x*x+x)/(x*x*x+1)+(x*x+x+1)/[(1-x*x)/x)]
当1<x<2时化简| x-1|/1-x-x/|x|-|x-2|/2-x
5x-2/(x+2)(x-2)-(x-1)(x+2)/(x+2)(x-2)=0
计算[(x-1)/(x-2)-(x-2)(x-3)-(x-3)/(x-4)+(x-4)(x-5)]/(2x-7)
已知x+1/x =4,则x^2/x^4+x^2+1
X^2-2X-1=0,X<0问X+1/X=?