Question

若m^2=m+1,n^2-n-1=0且m不等于n,求代数式m^7+n^7的值

由m^2=m+1,得，m^2-m-1=0,
又n^2-n-1=0，知，m,n是方程x^2-x-1=0的两根，（想问一下，前面这一步是怎么得出的？）

Answer 1

由m^2=m+1,得，m^2-m-1=0,
又n^2-n-1=0，知，m,n是方程x^2-x-1=0的两根，
由根与系数关系，得，m+n=1,mn=-1,
所以m^2+n^2
=(m+n)^2-2mn
=1+2
=3,
m^4+n^4
=(m^2+n^2)^2-2m^2n^2
=9-2
=7
又（m^2+n^2)(n^4+n^4)=m^6+m^2n^4+m^4n^2+n^6
即21=m^6+n^6+m^2n^2(m^2+n^2)
解得m^6+n^6=21-3=18
所以m^7+n^7
=(m+n)(m^6-m^5n+m^4n^2+m^3n^3-m^2n^4-mn^5+n^6)
=m^6-m^5n+m^4n^2+m^3n^3-m^2n^4-mn^5+n^6
=m^6+n^6+m^4-m^3n+m^2n^2-mn^3+n^4(mn=-1代入)
=m^6+n^6+m^4+n^4+m^2+n^2+1
=18+7+3+1
=29
以上回答你满意么？

追问

m,n是方程x^2-x-1=0的两根，这是怎么得到的？

Answer 2

由m^2=m+1,得，m^2-m-1=0,又n^2-n-1=0，知，(m,n不相等）m,n是方程x^2-x-1=0的两不同根，m+n=1.在不断利用x^2=x+1
x^7=x^5(x+1)=x^3(x+1)^2=x(x+1)(x^2+2X+1)=(x^2+x)(3x+2)=(2x+1)(3x+2)=6x^2+7X+2=6x+6+7x+2=13x+8
m^7+n^7=13(m+n)+16=13+16=29

Answer 3