16X^3-6X^2-4X+1=0 这道方程如何解呢,答案是多少呢

16X^3-6X^2-4X+1=0 这道方程如何解呢,答案是多少呢
看不懂啊,能不能把答案再说一遍,实数范围内就行了.
幻影2001 1年前 已收到1个回答 举报

ee市昌厦建筑 幼苗

共回答了16个问题采纳率:93.8% 举报

Mathematica求解结果;
In[11]:= Solve[16 x^3 - 6 x^2 - 4 x + 1 == 0,x]
Out[11]= {{x ->
1/8 + (-63 + 4 [ImaginaryI] Sqrt[1038])^(1/3)/(8 3^(2/3)) + 19/(
8 (3 (-63 + 4 [ImaginaryI] Sqrt[1038]))^(1/3))},{x ->
1/8 - ((1 + [ImaginaryI] Sqrt[3]) (-63 +
4 [ImaginaryI] Sqrt[1038])^(1/3))/(16 3^(2/3)) - (
19 (1 - [ImaginaryI] Sqrt[3]))/(
16 (3 (-63 + 4 [ImaginaryI] Sqrt[1038]))^(1/3))},{x ->
1/8 - ((1 - [ImaginaryI] Sqrt[3]) (-63 +
4 [ImaginaryI] Sqrt[1038])^(1/3))/(16 3^(2/3)) - (
19 (1 + [ImaginaryI] Sqrt[3]))/(
16 (3 (-63 + 4 [ImaginaryI] Sqrt[1038]))^(1/3))}}
[ImaginaryI]表示复数i

1年前

6
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