已知a的平方=a+1，求代数式a的五次方-5a+2的值。

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解决时间 2021-02-21 17:43

提问者网友：流星是天使的眼泪
2021-02-21 00:02

最佳答案

五星知识达人网友：一秋
2021-02-21 00:42

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

全部回答

1楼网友：雪起风沙痕
2021-02-21 07:38

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

2楼网友：廢物販賣機
2021-02-21 06:28

a^2=a+1
两边平方
a^4=a^2+2a+1=(a+1)+2a+1=3a+2
a^5=a*a^4=a(3a+2)=3a^2+2a=3(a+1)+2a=5a+2
所以a^5-5a+2=5a+2-5a+2=4

3楼网友：平生事
2021-02-21 04:53

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

4楼网友：等灯
2021-02-21 03:24

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

5楼网友：佘樂
2021-02-21 02:20

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

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