Question

已知m^2=m+1,n^2=n+1,m不等于n,求m^5+n^5的值

Answer 1

m^2=m+1,n^2=n+1,m不等于n
所以mn是方程x^2=x+1的两根，也就是x^2-x-1=0
由韦达定理m+n=1 mn=-1
所以m^2+n^2=(m+n)^2-2mn=1+2=3
m^3+n^3=(m+n)(m^2+n^2)-mn^2-nm^2=(m+n)(m^2+n^2)-mn(m+n)=1*3-1*(-1)=4
m^4+n^4=(m^2+n^2)^2-2m^2n^2=3^2-2*(-1)^2=9-2=7
m^5+n^5=(m+n)(m^4+n^4)-mn^4-nm^4=(m+n)(m^4+n^4)-mn(m^3+n^3)=1*7-(-1)*4=11

Answer 2

我思路出问题了 ，不好意思

Answer 3

易知,m和n是方程:x²-x-1=0的两个不相等的实根.
∴m+n=1,mn=-1.
[[1]]
m³+n³=(m+n)(m²-mn+n²)=(m+n)²-3mn=1+3=4
m³+n³=4
[[2]]
m²+n²=(m+n)²-2mn=1+2=3
m²+n²=3
[[3]]
12
=(m³+n³)(m²+n²)
=m^5+n^5+(mn)²(m+n)
=m^5+n^5+1
∴m^5+n^5=11