如何解一元三次方程？例如6X^3-7X^2+1

2024-11-27 18:05:19

推荐回答（4个）

回答1：

6X³-7X²+1
= (6x³-6x²)-(x²-1)
=6x²(x-1)-(x-1)(x+1)
=(x-1)(6x²-x-1)
=(x-1)(3x+1)(2x-1)=0
因此解为 1 -1/3 0.5
希望有帮助，不清楚请追问，有用请采纳 o(∩_∩)o

回答2：

6X^3-7X^2+1
=6x^3-6x^2-x^2+1
=6x^2(x-1)-(x^2-1)
=6x^2(x-1)-(x-1)(x+1)
=(x-1)(6x^2-x-1)=0
x1=1 x2=1/2 x3=-1/3

回答3：

6X^3-7X^2+1=0吧？

6X^3-6X^2 -x^2+1=0
6x^2(x-1)- (x^2-1)=0
6x^2(x-1)- (x-1)(x+1) =0
(x-1)(6x^2 -x-1)=0
（x-1)（2x-1)(3x+1)=0
x=1或x=1/2 或x=- 1/3

回答4：

可以用二分法求近似值