解答: e^(x-1)>x^n/n!在n=1时立 假充e^(x-1)>x^n/n!在n=k时成立 即e^(x-1) > x^k/k! e^(x-1) - x^k/k! >0 则当n=k+1时 z(x) = e^(x-1)-x^(k+1)/(k+1)! z1(x) = e^(x-1) - (k+1)x^k/(k+1)! = e^(x-1) - x^k/k!>0 由上一步n=k时的结论
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