> 数学 >
若3n^2-n=1,求6n^3+7n^2-5n+2003的值
人气:236 ℃ 时间:2019-10-09 10:53:54
解答
3n^2-n=1,
6n^3+7n^2-5n+2003
=6n^3-2n^2+9n^2-5n+2003
=2n(3n^2-n)+9n^2-5n+2003
=2n*1+9n^2-5n+2003
=9n^2-3n+2003
=3(3n^2-n)+2003
=3*1+2003
=2006
推荐
猜你喜欢
© 2024 79432.Com All Rights Reserved.
电脑版|手机版