初二的分式计算题:
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初二的分式计算题:
1/[a(a+1)]+1/[(a+1)(a+2)+1/[(a+2)(a+3)]+……+1/[(a+2004)(a+2005)]
=1/a-1/(a+1)+1/(a+1)-1/(a+2)+1/(a+2)-1/(a+3)+……+1/(a+2004)-1/(a+2005)
=1/a-1/(a+2005)
=2005/[a(a+2005)]
2.若分式 (3x-1)/[(x-1)(x-2)]=A/(x-1)+B/(x-2) ,那么,
A/(x-1)+B/(x-2)
=[A(x-2)+B(x-1)]/[(x-1)(x-2)]
=[x(A+B)-2A-B]/[(x-1)(x-2)]
则:A+B=3 …………(1)
2A+B=1…………(2)
(2)-(1)得:A=-2
代入(1)得:B=5 .
即:A=-2 B=5 .
再问: 谢谢了 其他的题目呢?
=1/a-1/(a+1)+1/(a+1)-1/(a+2)+1/(a+2)-1/(a+3)+……+1/(a+2004)-1/(a+2005)
=1/a-1/(a+2005)
=2005/[a(a+2005)]
2.若分式 (3x-1)/[(x-1)(x-2)]=A/(x-1)+B/(x-2) ,那么,
A/(x-1)+B/(x-2)
=[A(x-2)+B(x-1)]/[(x-1)(x-2)]
=[x(A+B)-2A-B]/[(x-1)(x-2)]
则:A+B=3 …………(1)
2A+B=1…………(2)
(2)-(1)得:A=-2
代入(1)得:B=5 .
即:A=-2 B=5 .
再问: 谢谢了 其他的题目呢?