Question

设x=根号5-1/2，求x4+x2+2x-l的值

Answer 1

X=(√5-1)/2，
2X+1=√5，
4X^2+4X+1=5，
X^2=1-X，降次法，
X^4+X^2+2X-1
=(1-X)^2+(1-X)+2X-1
=1-2X+X^2+X
=1-X+(1-X)
=2-2X
=2-(√5-1)
=3-√5。

Answer 2

　　
X=(√5-1)/2 x^2=(〖(√5-1)/2)〗^2=(5-2√5+1)/4 =(3-√5)/2
x^4+x^2+2x-1
=(x^2+1)( x^2-1)+x(x+2)
=((3-√5)/2+1)((3-√5)/2-1)+(√5-1)/2((√5-1)/2+2)
=(5-√5)/2×(1-√5)/2+(√5-1)/2×(√5+3)/2
=(√5-1)/2×(√5+3)/2-(5-√5)/2×(√5-1)/2
=(√5-1)/2×((√5+3)/2-(5-√5)/2)
=(√5-1)/2×(2√5-2)/2
=(√5-1)/2×(√5-1)
=(5-2√5+1)/2
=3-√5