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由f(x)=1-ex,得f(0)=1-e0=0.又f′(x)=-ex,∴f′(0)=-e0=-1.∴f(x)=1-ex在點P(0,0)處的切線方程為y-0=-1×(x-0),即x+y=0.故答案為:x+y=0.
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