设x=〔(√5)-1]/2,求x^4+x^2+2x-1的值
来源:百度知道 编辑:UC知道 时间:2024/09/27 05:43:35
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
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
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
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