观察下列关系式,1/(x-1)(x-2)={1/(x-2)}-{1/(x-1)}、.你可归纳出一般的结论是?计算1/(x-1)(x-2)+1/(x-2)(x-3)+.到1/(x-99)(x-100)

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观察下列关系式,1/(x-1)(x-2)={1/(x-2)}-{1/(x-1)}、.你可归纳出一般的结论是?计算1/(x-1)(x-2)+1/(x-2)(x-3)+.到1/(x-99)(x-100)
观察下列关系式,1/(x-1)(x-2)={1/(x-2)}-{1/(x-1)}、.你可归纳出一般的结论是?
计算1/(x-1)(x-2)+1/(x-2)(x-3)+.到1/(x-99)(x-100)

观察下列关系式,1/(x-1)(x-2)={1/(x-2)}-{1/(x-1)}、.你可归纳出一般的结论是?计算1/(x-1)(x-2)+1/(x-2)(x-3)+.到1/(x-99)(x-100)
可以归纳出一般的结论是:相邻的两个自然数积的倒数等于这两个自然数的倒数之差!
1/(x-1)(x-2)+1/(x-2)(x-3)+……+1/(x-99)(x-100)
=1/(x-1)-1/(x-2)+1/(x-2)-1/(x-3)+……+1/(x-99)-1/(x-100)
=1/(x-1)-1/(x-100)
=[(x-100)-(x-1)]/[(x-1)(x-100)]
=-99/[(x-1)(x-100)]
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您好:

1/(x-1)(x-2)
={1/(x-2)}-{1/(x-1)}

1/(x-1)(x-2)+1/(x-2)(x-3)+。。。。。到1/(x-99)(x-100)
=1/(x-2)-1/(x-1)+1/(x-3)-1/(x-2)+...+1/(x-100)-1/(x-99)
=1/(x-100)-1/(x-1)...

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您好:

1/(x-1)(x-2)
={1/(x-2)}-{1/(x-1)}

1/(x-1)(x-2)+1/(x-2)(x-3)+。。。。。到1/(x-99)(x-100)
=1/(x-2)-1/(x-1)+1/(x-3)-1/(x-2)+...+1/(x-100)-1/(x-99)
=1/(x-100)-1/(x-1)
=(x-1-x+100)/(x-1)(x-100)
=99/(x-1)(x-100)

不明白,可以追问
如有帮助,记得采纳,谢谢
祝学习进步!

收起

结果为1/(x-100)-1/(x-1)

1/(x-1)(x-2)+1/(x-2)(x-3)+……+1/(x-99)(x-100)
=1/(x-1)-1/(x-2)+1/(x-2)-1/(x-3)+……+1/(x-99)-1/(x-100)
=1/(x-1)-1/(x-100)
=[(x-100)-(x-1)]/[(x-1)(x-100)]
=-99/[(x-1)(x-100)]

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