解方程 x(x+1)(x²-2x-4)=0 x³-2x+1=0 x的4次方+8x³+14x²+8x+1=0
问题描述：
解方程 x(x+1)(x²-2x-4)=0 x³-2x+1=0 x的4次方+8x³+14x²+8x+1=0
答
x(x+1)(x²-2x-4)=0
x1=0 x2=-1
x³-2x+1=0
x³-x-x+1=0
x(x^2-1)-(x-1)=0
x(x+1)(x-1)-(x-1)=0
(x^2+x-1))(x-1)=0
x=1
x^4+8x³+14x²+8x+1=0
x^4+8x^3+7x^2+7x^2+8x+1=0
x^2(x^2+8x+7)+(7x+1)(x+1)=0
x^2(x+7)(x+1)+(7x+1)(x+1)=0
(x^3+7x^2+7x+1)(x+1)=0
{(x^3+1)+7x(x+1)}(x+1)=0
(x+1)(x^2-x+1+7x)(x+1)=0
(x^2+7x+1)(x+1)^2=0
x=-1
答
x(x+1)(x²-2x-4)=0 x1=0x2=-1x²-2x-4=0x²-2x+1=5(x-1)²=5x3=1+√5x4=1-√5x³-2x+1=0x³-x-x+1=0x(x²-1)-(x-1)=0x(x+1)(x-1)-(x-1)=0(x²+x-1))(x-1)=0x1=1x²+x-1=0x...