Question

设x=〔(√5）-1]/2，求x^4+x^2+2x-1的值

Answer 1

x=〔(√5）-1]/2
2x+1=√5
两边平方
4x^2+4x+1=5
4x^2+4x=4
x^2+x=1

x^4+x^2+2x-1
=x^4+x^3-x^3+x^2+2x-1
=x^2(x^2+x)-x^3+x^2+2x-1
=x^2*1-x^3+x^2+2x-1
=-x^3+2x^2+2x-1
=-x^3-x^2+3x^2+2x-1
=-x(x^2+x)+3x^2+2x-1
=-x*1+3x^2+2x-1
=3x^2+x-1
=3x^2+3x-2x-1
=3(x^2+x)-2x-1
=3*1-2x-1
=-2x+2
=-2*(√5-1)/2+2
=-√5+1+2
=3-√5

Answer 2

x^4+x^2+2x-1
=（x^4+2*x^2+1）-（x^2-2x+1）-1
=(x^2+1)^2-(x-1)^2-1
=(x^2+x)(x^2-x+2)-1

x^2=(3-根号5)/2
x=(根号5-1)/2
x^2+x=1

原式=x^2-x+1=3-根号5

Answer 3

x=〔(√5）-1]/2
则x^2=1，x^4=1，2x=√5-1
所以x^4+x^2+2x-1=√5
错了，不噶