已知四個實數
a
b
c
d,滿足
abcd=−5,
a(b−1)(c−1)(d−1)=11,
a(b−2)(c−2)(d−2)=33,
a(b−3)(c−3)(d−3)=73,則
a(b+1)(c+1)(d+1)的值為
。
[解答]
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