Question

m平方=m+1,n平方=n+1,且m不等于n ,求：m的7次方+n的7次方.请详细说明.

Answer 1

m^2-m-1=0,n^2-n-1=0

由题意知m,n为方程x^2-x-1=0的两个不同的实根.

由韦达定理知

m+n=1,mn=-1

m^2+n^2+2mn=(m+n)^2

m^2+n^2=1+2=3

(m^2+n^2)^2=m^4+2(mn)^2+n^4

9=m^4+n^4+2

m^4+n^4=7

(m^2+n^2)(m+n)=m^3+mn^2+m^2n+n^3

3=m^3+n^3+mn(m+n)

m^3+n^3=3-(-1)*1=4

(m^3+n^3)(m^4+n^4)=m^7+n^7+m^3n^4+m^4n^3=m^7+n^7+m^3n^3(m+n)

4*7=m^7+n^7+(-1)^3*1

m^7+n^7=28-(-1)^3*1=28+1=29