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  • X4+1因式分解

    X4+1因式分解
    百度之前查了的說(shuō)做不出來(lái),google也沒(méi)搜索到,這個(gè)怎么解?
    其他人氣:774 ℃時(shí)間:2020-04-16 16:04:28
    優(yōu)質(zhì)解答
    令 x⁴= -1 = -1 + i×0 = cos(2k+1)π + isin(2k+1)π = e^[i(2k+1)π]
    ∴ x = e^[¼i(2k+1)π]
    k=0,x.= e^[¼iπ] = cos(¼π) + isin(¼π) = √2/2 + i√2/2
    k=1,x₁= e^[¾iπ] = cos(¾π) + isin(¾π) = -√2/2 + i√2/2
    k=2,x₂= e^[5iπ/4] = cos(5π/4) + isin(5π/4) = -√2/2 - i√2/2
    k=3,x₃= e^[7iπ/4] = cos(7π/4) + isin(7π/4) = √2/2 - i√2/2
    k=4,x₄= e^[9iπ/4] = cos(9π/4) + isin(9π/4) = √2/2 + i√2/2
    .
    ∴ x⁴+ 1 = [x-(√2/2+i√2/2)][x-(-√2/2+i√2/2)][x-(-√2/2-i√2/2)][x-(√2/2-i√2/2)]
    = (x-√2/2-i√2/2)(x+√2/2-i√2/2)(x+√2/2+i√2/2)(x-√2/2+i√2/2)
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