x3-9x2+20x-3=0
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解决时间 2021-11-12 10:38
- 提问者网友：
- 2021-11-11 17:11
x3-9x2+20x-3=0
最佳答案
- 五星知识达人网友：鱼芗
- 2021-11-11 17:51
方程x^3-9x^2+20x-3=0的根为
X1=1/6*(-324+12*i*3387^(1/2))^(1/3)+14/(-324+12*i*3387^(1/2))^(1/3)+3
X2=-1/12*(-324+12*i*3387^(1/2))^(1/3)-7/(-324+12*i*3387^(1/2))^(1/3)+3+1/2*i*3^(1/2)*(1/6*(-324+12*i*3387^(1/2))^(1/3)-14/(-324+12*i*3387^(1/2))^(1/3))
X3=-1/12*(-324+12*i*3387^(1/2))^(1/3)-7/(-324+12*i*3387^(1/2))^(1/3)+3-1/2*i*3^(1/2)*(1/6*(-324+12*i*3387^(1/2))^(1/3)-14/(-324+12*i*3387^(1/2))^(1/3))
X1=1/6*(-324+12*i*3387^(1/2))^(1/3)+14/(-324+12*i*3387^(1/2))^(1/3)+3
X2=-1/12*(-324+12*i*3387^(1/2))^(1/3)-7/(-324+12*i*3387^(1/2))^(1/3)+3+1/2*i*3^(1/2)*(1/6*(-324+12*i*3387^(1/2))^(1/3)-14/(-324+12*i*3387^(1/2))^(1/3))
X3=-1/12*(-324+12*i*3387^(1/2))^(1/3)-7/(-324+12*i*3387^(1/2))^(1/3)+3-1/2*i*3^(1/2)*(1/6*(-324+12*i*3387^(1/2))^(1/3)-14/(-324+12*i*3387^(1/2))^(1/3))
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- 1楼网友：枭雄戏美人
- 2021-11-11 18:47
我把图片发给你😊追问好啊好啊
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