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)](/uploads/image/z/15096418-34-8.jpg?t=%E8%A7%82%E5%AF%9F%E4%B8%8B%E5%88%97%E5%85%B3%E7%B3%BB%E5%BC%8F%2C1%2F%28x-1%29%28x-2%29%3D%7B1%2F%28x-2%29%7D-%7B1%2F%28x-1%29%7D%E3%80%81.%E4%BD%A0%E5%8F%AF%E5%BD%92%E7%BA%B3%E5%87%BA%E4%B8%80%E8%88%AC%E7%9A%84%E7%BB%93%E8%AE%BA%E6%98%AF%3F%E8%AE%A1%E7%AE%971%2F%28x-1%29%28x-2%29%2B1%2F%EF%BC%88x-2%EF%BC%89%EF%BC%88x-3%EF%BC%89%2B.%E5%88%B01%2F%EF%BC%88x-99%EF%BC%89%EF%BC%88x-100%EF%BC%89)
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)]