∵ 1+2+3+···+n=n(n+1)/2, ∴ 1/(1+2+3+···+n) =2/[n(n+1)]=2[(1/n)-1/(n+1)], ∴ 原式=1+2[(1/2-1/3)+(1/3-1/4)+···+(1/2009-1/2010)] =1+2[(1/2)-(1/2010)] =1+1-(1/1005) =2-(1/1005) =2009/1005 =1+(1004/1005)(即带分数1又1005分之1004)
1+2+3+......+2007+2008+2009 =(1+2009)+(2+2008)+(3+2007)+......+(1004+1006)+1005 =2010*1004+1005 =2019045
答：答案不可能是楼主所说的x=9!理由如下: 当x=9时, 1/2*根(x+27)-1/2*根(13-x)+1/2*根x =1/2*根(9+27)-1/2*根(13...详情>>
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