高次方程求解X^4+8X-48=0

答案:2 悬赏:20 手机版

解决时间 2021-02-18 21:18

提问者网友：蔚蓝的太阳
2021-02-18 02:10

高次方程求解X^4+8X-48=0

最佳答案

五星知识达人网友：千夜
2021-02-18 03:27

X^4+8X-48=0解为x1=0.1443 + 2.6361ix2 = 0.1443 - 2.6361ix3 =2.4839x4= -2.7725======以下答案可供参考======供参考答案1:x1=1/2*2^(1/2)*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)+1/2*i*((2*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)*(4+4*257^(1/2))^(2/3)-32*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)+8*2^(1/2)*(4+4*257^(1/2))^(1/3))/(4+4*257^(1/2))^(1/3)/(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2))^(1/2)x2=1/2*2^(1/2)*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)-1/2*i*((2*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)*(4+4*257^(1/2))^(2/3)-32*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)+8*2^(1/2)*(4+4*257^(1/2))^(1/3))/(4+4*257^(1/2))^(1/3)/(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2))^(1/2)x3=-1/2*2^(1/2)*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)+1/2*((-2*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)*(4+4*257^(1/2))^(2/3)+32*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)+8*2^(1/2)*(4+4*257^(1/2))^(1/3))/(4+4*257^(1/2))^(1/3)/(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2))^(1/2)x4=-1/2*2^(1/2)*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)-1/2*((-2*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)*(4+4*257^(1/2))^(2/3)+32*(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2)+8*2^(1/2)*(4+4*257^(1/2))^(1/3))/(4+4*257^(1/2))^(1/3)/(((4+4*257^(1/2))^(2/3)-16)/(4+4*257^(1/2))^(1/3))^(1/2))^(1/2)

全部回答

1楼网友：第幾種人
2021-02-18 05:02

这个解释是对的

我要举报

如以上问答信息为低俗、色情、不良、暴力、侵权、涉及违法等信息，可以点下面链接进行举报！