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如何解一元三次方程式一般式?

学习 时间:2026-05-30 21:29:43 阅读：6573
如何解一元三次方程式一般式?知者请简洁道来,且莫有繁言琐语以扰人心扉
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留胡子的短靴

酷炫的书包

2026-05-30 21:29:43

卡当公式{{x -> -(b/(3 a)) - (2^(1/3) (-b^2 + 3 a c))/(3 a (-2 b^3 + 9 a b c - 27 a^2 d + Sqrt[4 (-b^2 + 3 a c)^3 + (-2 b^3 + 9 a b c - 27 a^2 d)^2])^(1/3)) + (-2 b^3 + 9 a b c - 27 a^2 d + Sqrt[4 (-b^2 + 3 a c)^3 + (-2 b^3 + 9 a b c - 27 a^2 d)^2])^(1/3)/(3 2^(1/3) a)},{x -> -(b/(3 a)) + ((1 + I Sqrt[3]) (-b^2 + 3 a c))/(3 2^(2/3)a (-2 b^3 + 9 a b c - 27 a^2 d + Sqrt[4 (-b^2 + 3 a c)^3 + (-2 b^3 + 9 a b c - 27 a^2 d)^2])^(1/3)) - (1/(6 2^(1/3)a))(1 - I Sqrt[3]) (-2 b^3 + 9 a b c - 27 a^2 d + Sqrt[4 (-b^2 + 3 a c)^3 + (-2 b^3 + 9 a b c - 27 a^2 d)^2])^(1/3)},{x -> -(b/(3 a)) + ((1 - I Sqrt[3]) (-b^2 + 3 a c))/(3 2^(2/3)a (-2 b^3 + 9 a b c - 27 a^2 d + Sqrt[4 (-b^2 + 3 a c)^3 + (-2 b^3 + 9 a b c - 27 a^2 d)^2])^(1/3)) - (1/(6 2^(1/3)a))(1 + I Sqrt[3]) (-2 b^3 + 9 a b c - 27 a^2 d + Sqrt[4 (-b^2 + 3 a c)^3 + (-2 b^3 + 9 a b c - 27 a^2 d)^2])^(1/3)}}

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  • 落后的鸭子
    回复
    2026-05-30 21:29:43

    卡当公式{{x -> -(b/(3 a)) - (2^(1/3) (-b^2 + 3 a c))/(3 a (-2 b^3 + 9 a b c - 27 a^2 d + Sqrt[4 (-b^2 + 3 a c)^3 + (-2 b^3 + 9 a b c - 27 a^2 d)^2])^(1/3)) + (-2 b^3 + 9 a b c - 27 a^2 d + Sqrt[4 (-b^2 + 3 a c)^3 + (-2 b^3 + 9 a b c - 27 a^2 d)^2])^(1/3)/(3 2^(1/3) a)},{x -> -(b/(3 a)) + ((1 + I Sqrt[3]) (-b^2 + 3 a c))/(3 2^(2/3)a (-2 b^3 + 9 a b c - 27 a^2 d + Sqrt[4 (-b^2 + 3 a c)^3 + (-2 b^3 + 9 a b c - 27 a^2 d)^2])^(1/3)) - (1/(6 2^(1/3)a))(1 - I Sqrt[3]) (-2 b^3 + 9 a b c - 27 a^2 d + Sqrt[4 (-b^2 + 3 a c)^3 + (-2 b^3 + 9 a b c - 27 a^2 d)^2])^(1/3)},{x -> -(b/(3 a)) + ((1 - I Sqrt[3]) (-b^2 + 3 a c))/(3 2^(2/3)a (-2 b^3 + 9 a b c - 27 a^2 d + Sqrt[4 (-b^2 + 3 a c)^3 + (-2 b^3 + 9 a b c - 27 a^2 d)^2])^(1/3)) - (1/(6 2^(1/3)a))(1 + I Sqrt[3]) (-2 b^3 + 9 a b c - 27 a^2 d + Sqrt[4 (-b^2 + 3 a c)^3 + (-2 b^3 + 9 a b c - 27 a^2 d)^2])^(1/3)}}

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