6.设ω=cos(2π/5) + i×sin(2π/5),则以ω,ω^3,ω^7,ω^9为根的方程是
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6.设ω=cos(2π/5) + i×sin(2π/5),则以ω,ω^3,ω^7,ω^9为根的方程是
(A) x4+x3+x2+x+1=0 (B) x4x3+x2x+1=0
(C) x4x3x2+x+1=0 (D) x4+x3+x2x1=0
答案我看不懂
(A) x4+x3+x2+x+1=0 (B) x4x3+x2x+1=0
(C) x4x3x2+x+1=0 (D) x4+x3+x2x1=0
答案我看不懂
这个答案都有搞笑,不要管他,哈哈
正确答案确实是B x4-x3+x2-x+1=0
证明如下;
ω=cos(2π/5) + i×sin(2π/5),
说明1,w,w^2,w^3,w^4是x^5-1的五个根
所以w^5=1,且(x-1)(x-w)(x-w^3)(x-w^2)(x-w^4)=x^5-1
=(x-1)(x4-x3+x2-x+1)
所以 (x-w)(x-w^3)(x-w^2)(x-w^4)
=(x^5-1)/(x-1)
=x4-x3+x2-x+1
w^7=w^2,w^9=w^4
所以(x-w)(x-w^3)(x-w^7)(x-w^9)
=(x-w)(x-w^3)(x-w^2)(x-w^4)
=x4-x3+x2-x+1
正确答案确实是B x4-x3+x2-x+1=0
证明如下;
ω=cos(2π/5) + i×sin(2π/5),
说明1,w,w^2,w^3,w^4是x^5-1的五个根
所以w^5=1,且(x-1)(x-w)(x-w^3)(x-w^2)(x-w^4)=x^5-1
=(x-1)(x4-x3+x2-x+1)
所以 (x-w)(x-w^3)(x-w^2)(x-w^4)
=(x^5-1)/(x-1)
=x4-x3+x2-x+1
w^7=w^2,w^9=w^4
所以(x-w)(x-w^3)(x-w^7)(x-w^9)
=(x-w)(x-w^3)(x-w^2)(x-w^4)
=x4-x3+x2-x+1
6.设ω=cos(2π/5) + i×sin(2π/5),则以ω,ω^3,ω^7,ω^9为根的方程是
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