X的三次减2X加1等于0,X等于多少
问题描述：
X的三次减2X加1等于0,X等于多少
答
= =、三元一次方程。。
答
x²-x²+x²-2x+1=0
x²(x-1)+(x-1)²=0
(x-1)(x²-x+x-1)=0
(x-1)²(x+1)=0
x=1,x=-1
答
解x³-2x+1=0x³-x²+x²-x-x+1=0(x³-x²)+(x²-x)-(x-1)=0x²(x-1)+x(x-1)-(x-1)=0(x-1)(x²+x-1)=0∴x-1=0或x²+x-1=0∴x=1或x=(-1±√5)/2
答
x^3-2x+1=0
x^3-x-(x-1)=0
x(x^2-1)-(x-1)=0
x(x-1)(x+1)-(x-1)=0
(x-1)(x^2+x-1)=0
x-1=0, x^2+x-1=0
x=1, x=(-1±√5)/2