16X^3-6X^2-4X+1=0 这道方程如何解呢
来源:百度知道 编辑:UC知道 时间:2024/07/03 02:14:30
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
1X 2X 3X 4X 16X 18X到底是什么意思?
x^6+x^5+x^4+x^3+x^2+x+1=0
1x 2x 3x 4x 5x 6=( )x( )+( )
x-1/x^2+3x+2+6/2+x-x^2-10-x/4-x^2
X*X-2X-1=0 求2x*x*x-3*x*x-4*x+2
解方程(x+1)(x+2)(x+3)(x+4)=(x+1)(x+1)+(x+2)(x+2)+(x+3)(x+3)+(x+4)(x+4)
(x+1)(x+2)(x+3)(x+4)=(x+1)(x+1)+(x+2)(x+2)+(x+3)(x+3)+(x+4)(x+4)
(x^2+x+1)(x^2-6x+1)+x^2 x^4+3x^3+x^2-3x-2
(1)(x^2+4)^2-16x^2 (2)(x+2)(x+3)+3x+1
因式分解(X+1)(X+2)(X+3)(X+4)+1