作业帮[1/(x-1)(x+2)]+[1/(x+2)(x+5)]+[1/(x+5)(x+8)]+[1/(x+8)(x+11)]=(1/3x-3)-1/24
来源：学生作业帮助网 编辑：作业帮 时间：2024/11/08 01:42:57
[1/(x-1)(x+2)]+[1/(x+2)(x+5)]+[1/(x+5)(x+8)]+[1/(x+8)(x+11)]=(1/3x-3)-1/24
[1/(x-1)(x+2)]+[1/(x+2)(x+5)]+[1/(x+5)(x+8)]+[1/(x+8)(x+11)]=(1/3x-3)-1/24
[1/(x-1)(x+2)]+[1/(x+2)(x+5)]+[1/(x+5)(x+8)]+[1/(x+8)(x+11)]=(1/3x-3)-1/24
[1/(x-1)(x+2)]+[1/(x+2)(x+5)]+[1/(x+5)(x+8)]+[1/(x+8)(x+11)]=(1/3x-3)-1/24
(1/3)[1/(x-1)-1/(x+11)]=1/[3(x-1)]-1/24
1/(x-1)-1/(x+11)=1/(x-1)-1/8
1/(x+11)=1/8
x=-3
证明：
左边=1/3[1/(x-1)-1/(x+2)+1/(x+2)-1/(x+5)+1/(x+5)-1/(x+8)+1/(x+8)-1/(x+11)]
=1/3[1/(x-1)-1/(x+11)]=4/(x-1)(x+11)
话说我只能证出这个结果，不是我题看的不对就是这道题有问题~~
1/(x-1)(x+2)=1/3{3/[(x-1)(x+2)]}=1/3{[(x+2)-(x-1)]/[(x-1)(x+2)]}
=1/3{(x+2)/[(x-1)(x+2)]-(x-1)]/[(x-1)(x+2)]}=1/3[1/(x-1)-1/(x+2)]
同理1/(x+2)(x+5)=1/3[1/(x+2)-1/(x+5)]
......
所以去括号抵消后[1...
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1/(x-1)(x+2)=1/3{3/[(x-1)(x+2)]}=1/3{[(x+2)-(x-1)]/[(x-1)(x+2)]}
=1/3{(x+2)/[(x-1)(x+2)]-(x-1)]/[(x-1)(x+2)]}=1/3[1/(x-1)-1/(x+2)]
同理1/(x+2)(x+5)=1/3[1/(x+2)-1/(x+5)]
......
所以去括号抵消后[1/(x-1)(x+2)]+[1/(x+2)(x+5)]+[1/(x+5)(x+8)]+[1/(x+8)(x+11)]=(1/3x-3)-1/24
就变成了(1/3)[1/(x-1)-1/(x+11)]=1/[3(x-1)]-1/24
......
收起