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更新1:
Factorize
MATHS F3 Factorize
回答 (3)
2010-11-03 6:18 am
✔ 最佳答案
12v - 12v^2 - 6u + 3u^2=12v - 6u - 12v^2 + 3u^2
=6(2v - u) - 3( 4v^2 - u^2)
=6(2v - u) - 3(2v+u)(2v-u)
=3(2v-u)[ 2 - (2v+u) ]
=3(2v-u)(2-2v-u)
2010-11-04 2:35 am
12v-12v^2-6u+3u^2
=12v-6u-(12v^2-3u^2)
=6(2v-u) - 3(4v^2-u^2)
=6(2v-u) - 3[(2v)^2-u^2]
=6(2v-u) - 3(2v+u)(2v-u)
= 3(2v-u) [2-(2v+u)]
=3(2v-u)(2-2v-u)
=12v-6u-(12v^2-3u^2)
=6(2v-u) - 3(4v^2-u^2)
=6(2v-u) - 3[(2v)^2-u^2]
=6(2v-u) - 3(2v+u)(2v-u)
= 3(2v-u) [2-(2v+u)]
=3(2v-u)(2-2v-u)
2010-11-03 4:33 pm
12v－12v^2－6u+3u^2
=(3u^2－6u+3)－(12v^2－12v+3)
=3(u^2－2u+1)－3(4v^2－4v+1)
=3(u－1)^2－3(2v－1)^2
=3(u－1－2v+1)(u－1+2v－1)
=3(u－2v)(u+2v－2)
=(3u^2－6u+3)－(12v^2－12v+3)
=3(u^2－2u+1)－3(4v^2－4v+1)
=3(u－1)^2－3(2v－1)^2
=3(u－1－2v+1)(u－1+2v－1)
=3(u－2v)(u+2v－2)
收錄日期: 2021-04-27 16:36:31
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20101102000051KK01614