Question

各位数学高手帮帮忙吧 已知x²－3x＋1＝0 求x²＋x的五次方分之一

Answer 1

由已知条件x²－3x＋1＝0 得：x²＋x=4x-1
(x^2＋x)^5
=(4x-1)^5
=(4x-1)^2 * (4x-1)^2 *(4x-1)
=(16x^2-8x+1) * (16x^2-8x+1) * (4x-1)
=[16(x^2+x)-24x+1] * [16(x^2+x)-24x+1] * (4x-1)
=[16(4x-1)-24x+1] * [16(4x-1)-24x+1] * (4x-1)
=[40x-15] * [40x-15] * (4x-1)
=25[8x-3] * [8x-3] * (4x-1)
=25[64x^2-48x+9] * (4x-1)
=25[64(x^2+x)-112x+9] * (4x-1)
=25[64(4x-1)-112x+9] * (4x-1)
=25[144x-55] * (4x-1)
......最后把x1，x2分别代入就得到x²＋x的五次方的值，再取倒数

Answer 2

（5+/-2根号5）^(-5)