16X^3-6X^2-4X+1=0 这道方程如何解呢,答案是多少呢
来源:学生作业帮 编辑:搜搜做题作业网作业帮 分类:数学作业 时间:2024/05/17 01:38:48
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.
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
16X^3-6X^2-4X+1=0 这道方程如何解呢,答案是多少呢
(7/x²+x)+(4/x²-x)=6/x²-1解分式方程.急呢~
解方程:(4/x-2)+(x-1/x-5x+6)+(2/3-x)=0
解方程:2x+4x+6x...+100x=1-(x+3x+5x+...+99x)
解方程 x+2/x+1+x+8/x+7=x+6/x+5+x+4/x+3
解方程x/(x-2)=2x/(x-3)+(1-x)/(x-5x+6)
x=6时,方程x-2=5成立吗?x=7呢?x=8呢?
请问 cos(x)^2/(cos(x)*sin(x)+a) = b 这个方程中的角x要如何求出呢?
几道初一数学题(x^2-1)/(x-3x) +( x-1)/(2x-1)=0 解方程已知方程 x分之3 + x-1分之6
解方程 :8*x(4)+16*x(3)+8*x(2)+x-1=0
解方程:x(6-x)-2x(x+1)-(x^3-3x^4)/x²=4
解分式方程:x-3/x-1-x-4/x-2=x-5/x-3-x-6