Question

已知a^2-a-1=0,求代数式a^5-5a+2的值

Answer 1

a^2-a-1=0
则
a^2=a+1

a^5-5a+2
=(a^2)*(a^3)-5a+2
=(a+1)*(a^3)-5a+2
=a^4+a^3-5a+2
=(a^2)*(a^2+a)-5a+2
=(a+1)(2a+1)-5a+2
=2a^2+3a+1-5a+2
=2a^2-2a+3
=2a+2-2a+3
=5

Answer 2

解：
∵a²-a-1=0
∴a²=a+1
∴a^5-5a+2
=a(a+1)²-5a+2
=a³+2a²-4a+2
=a(a+1)+2a²-4a+2
=3a²-3a+2
=3(a²-a)+2
=3+2
=5