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分析,要證明x>ln(1+x) 即是證明x-ln(1+x)>0 證明: 設t(x)=x-ln(1+x) 導數t'=1-1/(1+x) =x/(1+x)>0 所以,t是增函數, t(x)>t(0)=0 所以,x-ln(x+1)>0 囙此,x>ln(x+1)
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