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[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

分类: 作业答案 • 2022-01-17 23:50:36
问题描述：

[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

答

[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)]-1/241/(x-1)-1/(x+11)=1/(x-1)-1/81/(x+11)=1/8x=-3怎麼从[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变成(1/3)[1/(x-1)-1/(x+11)]=1/[3(x-1)]-1/24？[1/(x-1)(x+2)]+[1/(x+2)(x+5)]+[1/(x+5)(x+8)]+[1/(x+8)(x+11)] =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)]

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