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问题: ..2.(1)f(x)=(x^2+x+1)(x^2-x+1)(x^2-1)

2.(1)f(x)=(x^2+x+1)(x^2-x+1)(x^2-1)
(2)f(x)=(x+1)^2*(x-1)^2+2x^2
证明:
(1)==〉f[2^(1/6)]=1
(2)=/=〉f[2^(1/6)]=1

解答:

1)f(x)=(x^2+x+1)(x^2-x+1)(x^2-1)
=(x-1)(x^2+x+1)(x+1)(x^2-x+1)
=(x^3-1)(x^3+1)
=x^6-1
所以f(2^(1/6))=2-1=1.
2)f(x)=(x+1)^2*(x-1)^2+2x^2
=[(x+1)(x-1)]^2+2x^2
=(x^2-1)^2+2x^2
=x^4-2x^2+1+2x^2
=x^4+1
所以f(2^(1/6))=[2^(1/6)]^4+1=2^(2/3)+1=2*2^(1/3)+1<>1.