解三次方程:x三次方+x平方-5x+3=0,求x?

2025-03-09 11:58:30
推荐回答(4个)
回答1:

解:x三次方+x平方-5x+3=0
x^3+x^2-2x-(3x-3)=0
x(x^2+x-2)-3(x-1)=0
x(x+2)(x-1)-3(x-1)=0
(x-1)(x^2+2x-3)=0
(x-1)(x+3)(x-1)=0
所以,方程的根为:x1=x2=1,x3=-3

回答2:

解:(x^3)+x^2-5x+3=0

(x^3)+2x^2-3x-x^2-2x+3=0

x(x^2+2x-3)-(x^2+2x-3)=0

(x^2+2x-3)(x-1)=0

(x+3)(x-1)(x-1)=0

∴x[1]=-3,x[2]=x[3]=1

回答3:

x^3+x^2-5x+3=(x^3-x)+(x^2-4x+3)
=x(x+1)(x-1)+(x-1)(x-3)
=(x-1)(x^2+x+x-3)
=(X-1)(x^2+2x-3)
=(x-1)(x-1)(x+3)=0
x=1,-3

回答4:

X=1