计算1/[x(x+3)]+1/[(x+3)(x+6)]+.+1/[(x+15)(x+18)]=?1/3[(1/x+1/x+3)
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计算1/[x(x+3)]+1/[(x+3)(x+6)]+.+1/[(x+15)(x+18)]=?1/3[(1/x+1/x+3)计算1/[x(x+3)]+1/[(x+3)(x+6)]+.+1/[(x+1
计算1/[x(x+3)]+1/[(x+3)(x+6)]+.+1/[(x+15)(x+18)]=?1/3[(1/x+1/x+3)
计算1/[x(x+3)]+1/[(x+3)(x+6)]+.+1/[(x+15)(x+18)]=?
1/3[(1/x+1/x+3)
计算1/[x(x+3)]+1/[(x+3)(x+6)]+.+1/[(x+15)(x+18)]=?1/3[(1/x+1/x+3)
第一题.
原式=1/3 [ 1/x - 1/(x+3)+ 1/(x+3)- 1/(x+6)+.+ 1/(x+15) - 1/(x+18)
=1/3 [ 1/x - 1/(x+18) ]
=6/(x平方 + 18x)
原理:例如 1/[x(x+3)] = 1/x - 1/(x+3)
依此类推.可消去当中的项.
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