1) Factorize ( x + 1 ) ( x - 1 ) + { ( 1 - x ) / 3 }
2) Factorize - 1 0 p ^ 2 - 1 1 p q + 6 q ^ 2
3) Factorize { ( 3 m ^ 2 ) / 4 } - 3
F.2 Maths question
2007-11-17 7:10 pm
回答 (2)
2007-11-17 7:36 pm
✔ 最佳答案
1) ( x + 1 ) ( x - 1 ) + { ( 1 - x ) / 3 }= ( x^2 - 1 ) + { ( 1 - x ) / 3 }
= { 3 ( x^2 - 1 ) + ( 1 - x ) } / 3
= ( 3x^2 - x - 2 ) / 3
= { ( x - 1 ) ( 3x + 2 ) } / 3
2) -10p^2 - 11pq + 6q^2
= ( -5p + 2q ) ( 2p + 3q )
= - ( 5p - 2q ) ( 2p + 3q)
3) { ( 3m^2 ) / 4 } - 3
= ( 3m^2 - 12 ) / 4
= { 3 ( m^2 - 4 ) } / 4
= { 3 ( m + 2 ) ( m - 2 ) } / 4
參考: Myself
2007-11-17 7:44 pm
1) (x+1)(x-1) + [(1-x)/3]
= (x+1)(x-1) - [(x-1)/3]
= [(x-1)/3][3(x+1)-1]
= [(x-1)/3](3x+3-1)
= [(x-1)(3x+2)]/3
2)
3) [(3m^2)/4] - 3
= 3[(m^2)/4 - 1]
= 3[(m/2)^2 - (1)^2]
= 3[(m/2) + 1][(m/2) - 1]
= (x+1)(x-1) - [(x-1)/3]
= [(x-1)/3][3(x+1)-1]
= [(x-1)/3](3x+3-1)
= [(x-1)(3x+2)]/3
2)
3) [(3m^2)/4] - 3
= 3[(m^2)/4 - 1]
= 3[(m/2)^2 - (1)^2]
= 3[(m/2) + 1][(m/2) - 1]
收錄日期: 2021-04-13 22:25:24
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071117000051KK01071