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【简答题】
设函数f（x）=（x+a）lnx-x+a． （Ⅰ）设g（x）=f'（x），求g（x）函数的单调区间； （Ⅱ）若 a≥ 1 e ，试研究函数f（x）=（x+a）lnx-x+a的零点个数．
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举一反三
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