Question

解方程x4+2x3-6x2+2x+1=0

Answer 1

原式=(X4-2X2+1)+2*(X3-2X2+X)
=(X2-1)^2+2*X(X-1)^2
=(X-1)^2(X+1)^2+2*X(X-1)^2
=(X-1)^2((X+1)^2+2X)
=(X-1)^2(X2+4X+1)
所以X=1 或 X= 2+根号3 X=2-根号3

Answer 2

x^4+2x^3-6x^2+2x+1=0
x^4-x^3+3x^3-3X^2-(3x^2-3x)-(x-1)=0
x^3(x-1)+3x^2(x-1)-3x(x-1)-(x-1)=0
(x-1)(x^3+3x^2-3x-1)=0
(x-1)(x^3+x^2-x-1+2x^2-2x)=0
(x-1)[(x+1)^2(x-1)+2x(x-1)]=0
(x-1)^2[(x+1)^2+2x]=0
(x-1)^2(x^2+4x+1)=0
(x-1)^2[(x+2)^2-3]=0
得(x-1)^2=0
(x+2)^2-3=0

解得X1=X2=1
X3=√3-2 X4=-√3-2

Answer 3

解原式=(x-1)(x3+3x2-3x-1)=(x-1)2(x2+4x+1)

后面的你应该会了吧

追问

这。。。我也找到了

Answer 4

(x^4-2x²+1)+(2x³-4x²+2x)=0
(x²-1)²+2x(x²-2x+1)=0
(x+1)²(x-1)²+2x(x-1)²=0
(x-1)²(x²+2x+1+2x)=0
(x-1)²(x²+4x+1)=0

x=1,x=-2-√3,x=-2+√3