请根据等式(b-a)/(ab)=b/(ab)-a/(ab)=1/a-1/b将分式1/[(x+1)]化为两个分式之差的形式.用上述规律,计算:1/[x(x+1)]1/[(x+1)(x+2)]+……+1/[(x+2007)(x+2008)]和:1/[x(x+3)]1/[(x+3)(x+6)]+……+1/[(x+27)(x+30)]
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![请根据等式(b-a)/(ab)=b/(ab)-a/(ab)=1/a-1/b将分式1/[(x+1)]化为两个分式之差的形式.用上述规律,计算:1/[x(x+1)]1/[(x+1)(x+2)]+……+1/[(x+2007)(x+2008)]和:1/[x(x+3)]1/[(x+3)(x+6)]+……+1/[(x+27)(x+30)]](/uploads/image/z/5153246-62-6.jpg?t=%E8%AF%B7%E6%A0%B9%E6%8D%AE%E7%AD%89%E5%BC%8F%28b-a%29%2F%28ab%29%3Db%2F%28ab%29-a%2F%28ab%29%3D1%2Fa-1%2Fb%E5%B0%86%E5%88%86%E5%BC%8F1%2F%5B%28x%2B1%29%5D%E5%8C%96%E4%B8%BA%E4%B8%A4%E4%B8%AA%E5%88%86%E5%BC%8F%E4%B9%8B%E5%B7%AE%E7%9A%84%E5%BD%A2%E5%BC%8F.%E7%94%A8%E4%B8%8A%E8%BF%B0%E8%A7%84%E5%BE%8B%2C%E8%AE%A1%E7%AE%97%EF%BC%9A1%2F%5Bx%28x%2B1%29%5D1%2F%5B%28x%2B1%29%28x%2B2%29%5D%2B%E2%80%A6%E2%80%A6%2B1%2F%5B%28x%2B2007%29%28x%2B2008%29%5D%E5%92%8C%EF%BC%9A1%2F%5Bx%28x%2B3%29%5D1%2F%5B%28x%2B3%29%28x%2B6%29%5D%2B%E2%80%A6%E2%80%A6%2B1%2F%5B%28x%2B27%29%28x%2B30%29%5D)
请根据等式(b-a)/(ab)=b/(ab)-a/(ab)=1/a-1/b将分式1/[(x+1)]化为两个分式之差的形式.用上述规律,计算:1/[x(x+1)]1/[(x+1)(x+2)]+……+1/[(x+2007)(x+2008)]和:1/[x(x+3)]1/[(x+3)(x+6)]+……+1/[(x+27)(x+30)]
请根据等式(b-a)/(ab)=b/(ab)-a/(ab)=1/a-1/b将分式1/[(x+1)]化为两个分式之差的形式.
用上述规律,计算:
1/[x(x+1)]1/[(x+1)(x+2)]+……+1/[(x+2007)(x+2008)]
和:
1/[x(x+3)]1/[(x+3)(x+6)]+……+1/[(x+27)(x+30)]
请根据等式(b-a)/(ab)=b/(ab)-a/(ab)=1/a-1/b将分式1/[(x+1)]化为两个分式之差的形式.用上述规律,计算:1/[x(x+1)]1/[(x+1)(x+2)]+……+1/[(x+2007)(x+2008)]和:1/[x(x+3)]1/[(x+3)(x+6)]+……+1/[(x+27)(x+30)]
1/[x(x+1)]
=[(x+1)-x]/[x(x+1)]
=(x+1)/[x(x+1)]-x/[x(x+1)]
=1/x-1/(x+1)
1/[x(x+1)]1/[(x+1)(x+2)]+……+1/[(x+2007)(x+2008)]
=1/x-1/(x+1)-1/(x+1)-1/(x-2)+……+1/(x+2007)-1/(x+2008)
=1/x-1/(x+2008)
=2008/[x(x+2008)]
1/[x(x+3)]+1/[(x+3)(x+6)]+……+1/[(x+27)(x+30)]
=(1/3){3/[x(x+3)]+3/[(x+3)(x+6)]+……+3/[(x+27)(x+30)]}
=(1/3)[1/x-1/(x+3)+1/(x+3)-1/(x+6)+……+1/(x+27)-1/(x+30)]
=(1/3)[1/x-1/(x+30)]
=10/[x(x+30)]
1/[x(x+1)]+1/[(x+1)(x+2)]+……+1/[(x+2007)(x+2008)]
=[1/x-1/(x+1)]+[1/(x+1)-1/(x+2)]+...+[1/(x+2007)-1/(x+2008)]
=1/x-1/(x+2008)
=2008/x(x+2008)
1/[x(x+3)]+1/[(x+3)(x+6)]+……+1/[(x+27)(x+...
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1/[x(x+1)]+1/[(x+1)(x+2)]+……+1/[(x+2007)(x+2008)]
=[1/x-1/(x+1)]+[1/(x+1)-1/(x+2)]+...+[1/(x+2007)-1/(x+2008)]
=1/x-1/(x+2008)
=2008/x(x+2008)
1/[x(x+3)]+1/[(x+3)(x+6)]+……+1/[(x+27)(x+30)]
=[1/x-1/(x+3)/3+[1/(x+3)-1/(x+6)]/3+...+[1/(x+27)-1/(x+30)]/3
=[1/x-1/(x+30)]/3
=[30/x(x+30)]/3
=10/x(x+30)
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