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函數f(x)的定義域為(0,+∞),函數的導數為f′(x)=lnx+x•1x=1+lnx,則由f(x0)+f′(x0)=1,即1+lnx0+xlnx0=1,得(x0+1)lnx0=0,解得x0=1或x0=-1(舍去),故x0=1,故答案為:1
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