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填充9.

設 f(x)=x^3+px^2+qx+5/a=0 的三根為b,c,d  , 此時bcd=-5/a,它符合了abcd=-5
則 f(x+1)=0  三根為b-1,c-1,d-1 =>(b-1)(c-1)(d-1)= -(1+p+q+5/a)=11/a => p+q=-1-16/a
則 f(x+2)=0  三根為b-2,c-2,d-2 =>(b-2)(c-2)(d-2)= -(8+4p+2q+5/a)=33/a => 2p+q=-4-19/a
則 f(x+3)=0  三根為b-3,c-3,d-3 =>(b-3)(c-3)(d-3)= -(27+9p+3q+5/a)=73/a => 3p+q=-9-26/a
可得 p=-3-3/a=-5-7/a => a=-2 , p=-3/2 , q=17/2
  f(x-1)=0  三根為b+1,c+1,d+1 => (b+1)(c+1)(d+1)=-(-1+p-q+5/a)=27/2
=> a(b+1)(c+1)(d+1)=-27

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