Question

已知x²=x+1，y²=y+1，且x≠y。求①x+y=1 ②x的五次方+y的五次方的值。

Answer 1

1、证明：由x²-x-1=0及y²-y-1=0可得：x=(1±√ 5)/2.y=(1±√5)/2而x≠y，所以x=(1+√5)/2,y=(1-√5)/2或x=（1-√5)/2, y=(1+√5)/2.分别把上面两个结果相加都可得：x+y=1，如x1+y1=(1+√5)/2+(1-√5）/2=1。
2、x^5+y^5=(x-y)(x^4+x^3y+x^2y²+xy³+y^4)=√5(x^4+x^3y+x^2y2+xy^3+y^4)=√5[(x²+y²+xy)(x²+y²-xy)+xy(x²+y²)]=√5[(x+y+2+xy)(x+y+2-xy)+(x+y)(x+y+2)]=√5[(3+xy)(3-xy)+3]=√5[12-x²y²]=√5(12-2-xy)=√5(10-xy)=√5(10+1)=11√5.