u^2 - 16v^2 - 4u - 16v
(a) a^4 + 2a^2b^2 + b^4
(b) a^4 + a^2b^2 + b^4
Factorization (急!!!!!!)
2010-09-21 6:47 am
回答 (1)
2010-09-21 7:05 am
✔ 最佳答案
u^2 - 16v^2 - 4u - 16v= (u^2 - (4v)^2) - (4u + 16v)= (u - 4v)(u + 4v) - 4(u + 4v)= (u + 4v)(u - 4v - 4)a)a^4 + 2a^2b^2 + b^4= (a^2)^2 + 2(a^2)(b^2) + (b^2)^2= (a^2 + b^2)^2b)a^4 + a^2b^2 + b^4= a^4 + 2(a^2)b^2 + b^4 - (a^2)b^2= (a^2 + b^2)^2 - (ab)^2= (a^2 + b^2 - ab)(a^2 + b^2 + ab)
收錄日期: 2021-04-21 22:16:38
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100920000051KK01651