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
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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](/uploads/image/z/3900170-2-0.jpg?t=6%EF%BC%8E%E8%AE%BE%CF%89%3Dcos%282%CF%80%2F5%29+%2B+i%C3%97sin%282%CF%80%2F5%29%2C%E5%88%99%E4%BB%A5%CF%89%2C%CF%89%5E3%2C%CF%89%5E7%2C%CF%89%5E9%E4%B8%BA%E6%A0%B9%E7%9A%84%E6%96%B9%E7%A8%8B%E6%98%AF%28A%29+x4%2Bx3%2Bx2%2Bx%2B1%3D0+%28B%29+x4%26%2361485%3Bx3%2Bx2%26%2361485%3Bx%2B1%3D0%28C%29+x4%26%2361485%3Bx3%26%2361485%3Bx2%2Bx%2B1%3D0+%28D%29+x4%2Bx3%2Bx2%26%2361485%3Bx%26%2361485%3B1%3D0%E7%AD%94%E6%A1%88%E6%88%91%E7%9C%8B%E4%B8%8D%E6%87%82)
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
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
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
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
B
把ω看作ω^1 1+9=3+7
所以ω1+ω3+ω7+ω9=1
结合韦达定理