题目
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.
看不懂啊,能不能把答案再说一遍,实数范围内就行了.
提问时间：2021-03-20
答案
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
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
举一反三
已知函数f（x）=x，g（x）=alnx，a∈R．若曲线y=f（x）与曲线y=g（x）相交，且在交点处有相同的切线，求a的值和该切线方程．
我想写一篇关于奥巴马的演讲的文章,写哪一篇好呢?为什么好
奥巴马演讲不用看稿子.为什么中国领导演讲要看?
想找英语初三上学期的首字母填空练习……
英语翻译
最新试题
热门考点