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设函数F(x)=13x3?A2x2+Bx+C,曲线y=F(x)在点(0,F(0))处的切线方程为y=...

(1)f'(x)=x2-ax+b,由题意,得f(0)=1f′(0)=0即c=1b=0.;(2

(1)f′(x)=x2-ax+b.由题意得f(0)=1f′(0)=0,即c=1b=0.所以b=0,c

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