Question

已知a的平方=a+1，求代数式a的五次方-5a+2的值。

Answer 1

因为 a*a=a+1
a的四次方=（a+1）的平方 = a*a +2a+1
a的五次方=a*a*a+2a*a+a ①
将①带入所求的方程
a的五次方-5a+2=a*a*a+2a*a-4a+2 ②
a的三次方又可以用已知条件替换
a*a*a=a*(a+1)
故②式=3a*a-3a +2
又因为a*a=a+1 （已知条件）
a*a-a=1
②式= 3+2=5

Answer 2

a²=a+2 采用逐渐降次法
a^5-5a+2
=a²*a³-5a+2
=(a+1)²*a-5a+2
=(a²+2a+1)*a-5a+2
=(3a+2)*a-5a+2
=3a²-3a+2
=3a+3-3a+2
=5

Answer 3

a的平方=a+1
a的平方-a-1=0
a的五次方-5a+2
=a^5-a^4-a^3+a^4-a^3-a^2+2a^3-2a^2-2a+3a^2-3a-3+5
=a^3(a^2-a-1)+a^2(a^2-a-1)+a(a^2-a-1)+(a^2-a-1)+5
=a^3*0+a^2*0+a*0+5
=5

Answer 4

因为a²-a=1
所以a^5-5a+2
=a^5-a^4+a^4-5a+2
=a^3(a^2-a)+a^4-5a+2
=a^3+(a^4-a^3)+a^3-5a+2
=2a^3+a^2-5a+2
=(2a^3-2a^2）+3a^2-5a+2
=2a+3a^2-5a+2
=3a^2-3a+2
=3×1+2
=5

Answer 5

a²=a+2 采用逐渐降次法
a^5-5a+2
=a²*a³-5a+2
=(a+1)²*a-5a+2
=(a²+2a+1)*a-5a+2
=(3a+2)*a-5a+2
=3a²-3a+2
=3a+3-3a+2
=5