Question

初一数学题 急急急急急 在线等

已知a^2=a+1,求代数式a^5-5a+2的值

Answer 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
这是我在静心思考后得出的结论，如果能帮助到您，希望您不吝赐我一采纳，如果不能请追问，我会尽全力帮您解决的

Answer 2

a^2=a+1,所以：a4次方=a²+2a+1=3a+2
原式=a(a4次方-5）+2=a(3a+2-5)+2=3a²-3a+2=3(a+1)-3a+2=5

Answer 3

你好

a^2=a+1
a^4=a^2+2a+1=a+1+2a+1=3a+2
a^5-5a+2
=a*a^4-5a+2
=a（3a+2）-5a+2
=3a²+2a-5a+2
=3a²-3a+2
=3（a+1）-3a+2
=5

~~

Answer 4

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

Answer 5

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