Question

已知a^2+a-1=0,求代数式a^5-5a+3的值

请用初中一年级的知识解答，谢谢您了！

Answer 1

a^5-5a+3

=a³﹙a²＋a-1﹚－a^4＋a³-5a+3
=a³﹙a²＋a-1﹚－a²﹙a²＋a-1﹚＋2a³-a²-5a+3
=2a﹙a²＋a-1﹚-3a²-3a+3
=-3﹙a²＋a-1﹚
=0

追问

a^5-5a+3

=a³﹙a²＋a-1﹚－a^4＋a³-5a+3
请问这是怎么得出来的？

追答

a^5-5a+3
=a^5+a^4-a³－a^4＋a³-5a+3

=a³﹙a²＋a-1﹚－a^4＋a³-5a+3