求a=1999时,3a³-2a²+4a-1-(3a³-3a²+3a-2001)的值

分类: 作业答案 • 2021-12-20 01:36:28

问题描述：

答

3a³-2a²+4a-1-(3a³-3a²+3a-2001)
=a²+a+2000
=1999²+1999+2000
=(2000-1)²+3999
=4000000-4000+1+3999
=4000000

答

3a³-2a²+4a-1-(3a³-3a²+3a-2001)=3a³-2a²+4a-1-3a³+3a²-3a+2001=a²+a+2000=a(a+1)+2000=1999×2000+2000=2000×(1999+1)=2000×2000=4000000