Question

若a²=a+1,b²=b+1,a≠b，求a^5+b^5的值

不要用韦达定理

Answer 1

有已知得：a,b就是方程x^2-x-1=0的两根，由韦达定理有：
a+b=1,ab=-1
a^5+b^5=(a+b)^5-5(a^4)b-10(a^3)(b^2)-10(a^2)(b^3)-5a(b^4)
=1+5a^3-10a-10b+5b^3
=1+5(a^3+b^3)-10(a+b)
=5(a+b)(a^2-ab+b^2)-9
=5(a^2+b^2+1)-9
=5((a+b)^2-2ab+1)-9
=11

Answer 2

a²=a+1
b²=b+1
相减的
(a+b)(a-b)=a-b
因为 a≠0
所以 a+b=0

a²=a+1,
a^4=(a+1)²=a²+2a+1

b²=b+1
b^4=(b+1)²=b²+2b+1

a^5+b^5
=a*a^4+b*b^4
=a(a²+2a+1)+b(b²+2b+1)
=a(a+1+2a+1)+b(b+1+2b+1)
=a(3a+2)+b(3b+2)
=3a²+2a+3b²+2b
=3(a+1)+2a+3(b+1)+2b
=5a+3+5b+3
=5(a+b)+6
=6

Answer 3

1