pa-pb-qa+qb,,plx factorize....點計?
回答 (7)
2006-11-06 6:04 pm
✔ 最佳答案
併項法 (grouping terms method)：把一個多項式分成數組，並於每組中各項進行提取公因式，然後再提取各組的公因式，從而把多項式分解為因式。
pa﹣pb﹣qa+qb
=p(a﹣b)﹣q(a﹣b)
=(a﹣b)(p﹣q)
或
pa﹣pb﹣qa+qb
=pa﹣qa﹣pb+qb
=a(p﹣q)﹣b(p﹣q)
=(p﹣q)(a﹣b)
參考: 中二級數學書
2006-11-07 3:56 am
pa-pb-qa+qb
=pa-qa-pb+qb
=a(p-q)+b(-p+q)
=a(p-q)-b(p-q)
=(a-b)(p-q)
=pa-qa-pb+qb
=a(p-q)+b(-p+q)
=a(p-q)-b(p-q)
=(a-b)(p-q)
2006-11-07 2:35 am
pa-pb-qa+qb
=pa-pb-(qa-qb)
=p(a-b)-q(a-b)
=(p-q)(a-b)
=pa-pb-(qa-qb)
=p(a-b)-q(a-b)
=(p-q)(a-b)
2006-11-06 7:52 pm
(pa - pb) - (qa + qb)
= p(a - b) - q(a - b)
= (p - q)(a - b)
= p(a - b) - q(a - b)
= (p - q)(a - b)
2006-11-06 5:47 pm
pa-pb-qa+qb
=p(a-b)-q(a-b)
=(a-b)(p-q)
=p(a-b)-q(a-b)
=(a-b)(p-q)
2006-11-06 5:46 pm
=p(a-b) - q (a-b)
=(p-q)(a-b)
=(p-q)(a-b)
收錄日期: 2021-04-12 22:26:43
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061106000051KK00716