F2的因式分解問題
回答 (3)
2010-11-24 5:11 am
✔ 最佳答案
16a^4 - 1= (4a^2)^2-1
= (4a^2+1) (4a^2-1)
= (4a^2+1) [(2a)^2-1]
= (4a^2+1) (2a+1) (2a-1)
參考: girl
2010-11-24 2:59 am
16a^4-1
= (4a^2)^2 - (1)^2
= (4a^2 + 1)(4a^2 - 1)
= (4a^2 + 1)(2a+1)(2a-1)
Hope can help you!
= (4a^2)^2 - (1)^2
= (4a^2 + 1)(4a^2 - 1)
= (4a^2 + 1)(2a+1)(2a-1)
Hope can help you!
參考: myself!
2010-11-24 2:35 am
因式分解 16a^4-1A. 16a^4-1= 2^4a^4-1^4= (2^2a^2-1^2)(2^2a^2+1^2)= (2a-1)(2a+1)(4a^2+1)
2010-11-23 18:36:02 補充：
因式分解 16a^4-1
A. 16a^4-1
= 2^4a^4-1^4
= (2^2a^2-1^2)(2^2a^2+1^2)
= (2a-1)(2a+1)(4a^2+1)
2010-11-23 18:36:02 補充：
因式分解 16a^4-1
A. 16a^4-1
= 2^4a^4-1^4
= (2^2a^2-1^2)(2^2a^2+1^2)
= (2a-1)(2a+1)(4a^2+1)
參考: Myself
收錄日期: 2021-04-23 23:13:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101123000051KK00922