a^2=a+1,b^2=b+1,a\=b,a^5+b^5=

a^2-a-1=0
b^2-b-1=0
a不等于b
所以a和b是方程x^2-x-1=0的两个根
所以a+b=1,ab=-1
a^2+b^2=(a+b)^2-2ab=1-2*(-1)=3
a^3+b^3=(a+b)(a^2-ab+b^2)=1*[3-(-1)]=4

a^5+b^5
=(a^2+b^2)(a^3+b^3)-a^2b^3-a^3b^2
=3*4-a^2b^2(a+b)
=12-(ab)^2(a+b)
=12-1*1
=11

x^2-x-1=0
a b为方程两根
a+b=1 ab=-1
a^2+b^2=(a+b)^2-2ab=3
a^4+b^4=(a^2+b^2)^2-2a^2*b^2=7
(a^4+b^4)(a+b)=ab(a^3+b^3)+a^5+b^5
a^3+b^3=(a+b)(a^2-ab+b^2)=4
所以a^5+b^5=7+4=11