中二因式分解...急,請幫幫手!!!!!20分
回答 (5)
2010-10-25 1:54 am
✔ 最佳答案
See this :9-9y²-4x²+12xy
= 3² - 9y² - 4x² + 12xy
= 3² - 9y² - ( +4x² ) - ( -12xy ) <--- 負負 = 正 , 負正 = 負
= 3² - ( 9y² + 4x² - 12xy )
= 3² - ( 3y + 2x ) ²
= [ 3 + ( 3y + 2x ) ] [ 3 - ( 3y + 2x ) ] <--- a² - b² = ( a + b ) ( a - b )
= ( 3 + 3y + 2x ) ( 3 - 3y - 2x )
參考: Hope I Can Help You ^_^
2010-10-30 11:42 pm
9-9y²-4x²+12xy
=9(1-y^2)+4x(-x+3y)
=(9+4x)(1-y^2)(-x+3y)
=(9+4x)(1-y)^2(-x+3y)
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唔係-4x²咩????點解負同正轉左既?
負同正轉左唔影響最終答案
=9(1-y^2)+4x(-x+3y)
=(9+4x)(1-y^2)(-x+3y)
=(9+4x)(1-y)^2(-x+3y)
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唔係-4x²咩????點解負同正轉左既?
負同正轉左唔影響最終答案
參考: me!!!
2010-10-25 2:34 am
9-9y^2-4x^2+12xy
=9-(3y)^2+12xy-(2x)^2
= 3^2-[(3y)^2-12xy+(2x)^2]
= 3^2-(3y-2x)^2
=[3+(3y-2x)][3-(3y-2x)]
=(3+3y-2x)(3-3y+2x)
=9-(3y)^2+12xy-(2x)^2
= 3^2-[(3y)^2-12xy+(2x)^2]
= 3^2-(3y-2x)^2
=[3+(3y-2x)][3-(3y-2x)]
=(3+3y-2x)(3-3y+2x)
2010-10-25 1:25 am
你條問題都幾搞笑!!
-9y²-4x²+12xy---------------------(factorize this question)
-(9y^2-12y+4x^2)
-(3y-2x)^2
-9y²-4x²+12xy---------------------(factorize this question)
-(9y^2-12y+4x^2)
-(3y-2x)^2
2010-10-25 12:53 am
9-9y²-4x²+12xy
=9-(9y²-12xy+4x²)
=3²-(3y-2x)²
=[3+(3y-2x)][3-(3y-2x
=(3+3y-2x)(3-3y+2x)
=9-(9y²-12xy+4x²)
=3²-(3y-2x)²
=[3+(3y-2x)][3-(3y-2x
=(3+3y-2x)(3-3y+2x)
參考: me
收錄日期: 2021-04-13 17:35:38
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https://hk.answers.yahoo.com/question/index?qid=20101024000051KK00984