(1) －121＋x^2 y^2
(2) 27a^2 b^2－3b^2
(3) 81(2h－k)^2－121(h－k)^2
(4) 81y^4－16z^4
更新1:
No. 3 wrong
更新2:
已知答案: 第一題: (xy+11)(xy-11) 第二題: 3b^2 (3a+1)(3a-1) 第三題: (7h+2k)(29h-20k) 第四題: (9y^2 + 4z^2)(3y+2z)(3y-2z) 只求步驟，謝謝。
[F.3數學] Factorization (40分)
2010-09-02 11:20 pm
Factorize the following expressions.
(1) －121＋x^2 y^2
(2) 27a^2 b^2－3b^2
(3) 81(2h－k)^2－121(h－k)^2
(4) 81y^4－16z^4
(1) －121＋x^2 y^2
(2) 27a^2 b^2－3b^2
(3) 81(2h－k)^2－121(h－k)^2
(4) 81y^4－16z^4
回答 (3)
2010-09-03 3:27 am
✔ 最佳答案
(1) －121＋x^2 y^2= (xy)^2 - (11)^2
= (xy-11)(xy+11)
(2) 27a^2 b^2－3b^2
= 3b^2 (9a^2 - 1)
= 3b^2 [(3a)^2 - 1^2]
= 3b^2 (3a-1)(3a+1)
(3) 81(2h－k)^2－121(h－k)^2
= [9(2h-k)]^2 - [11(h-k)]^2
= [9(2h-k)-11(h-k)][9(2h-k)+11(h-k)]
= (18h-9k-11h+11k)(18h-9k+11h-11k)
= (2k+7h)(29h-20k)
(4) 81y^4－16z^4
= (9y^2)^2 - (4z^2)^2
= (9y^2 - 4z^2)(9y^2 + 4z^2)
= [(3y)^2 - (2z)^2](9y^2 + 4z^2)
= (3y-2z)(3y+2z)(9y^2 + 4z^2)
參考: Hope the soluton can help you^^”
2010-09-03 4:15 am
81(2h－k)^2－121(h－k)^2
(18h-9k)^2-(11h-11k)^2
(18h-9k-11h+11k)(18h-9k+11h-11k)
(7h+2k)(29h-20k)
(18h-9k)^2-(11h-11k)^2
(18h-9k-11h+11k)(18h-9k+11h-11k)
(7h+2k)(29h-20k)
2010-09-02 11:44 pm
(a+b)(a-b)=a^2-b^2
(1)Exp.=x^2y^2-121
=(xy)^2-(11)^2
=(xy+11)(xy-11)
(2)Exp.=3(9a^2b^2-b^2)
=3((3ab)^2-(b)^2)
=3(3ab+b)(3ab-b)
=3b(3a+1)(3a-1)
(3)Exp.=(9(2h-k))^2-(11(2h-k))^2
=(20(2h-k))(-2(2h-k))
(4)Exp.=(9y^2)^2-(4z^2)^2
=(9y^2+4z^2)(9y^2-4z^2)
=(9y^2+4z^2)(3y+2z)(3y-2z)
(1)Exp.=x^2y^2-121
=(xy)^2-(11)^2
=(xy+11)(xy-11)
(2)Exp.=3(9a^2b^2-b^2)
=3((3ab)^2-(b)^2)
=3(3ab+b)(3ab-b)
=3b(3a+1)(3a-1)
(3)Exp.=(9(2h-k))^2-(11(2h-k))^2
=(20(2h-k))(-2(2h-k))
(4)Exp.=(9y^2)^2-(4z^2)^2
=(9y^2+4z^2)(9y^2-4z^2)
=(9y^2+4z^2)(3y+2z)(3y-2z)
收錄日期: 2021-04-24 10:31:07
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100902000051KK00673