1.) 6p^2 - 25pq + 14q^2
2.) 12c^2+40cd+25d^2
factorize 問題..please help help me!!
2006-12-14 5:50 am
回答 (4)
2006-12-14 5:54 am
✔ 最佳答案
1.) 6p^2 - 25pq + 14q^2=(2p-7q)(3p-2q)
2.) 12c^2+40cd+25d^2
=(6c+5d)(2c+5d)
2006-12-14 7:26 pm
1.) 6p^2-25pq+14q^2
=(2p)(3p)-25pq+(-7q)(-2q)................................a^2-2ab+b^2=(a-b)(a-b)
=(2p-7q)(3p-2q)
2.) 12c^2+40cd+25d^2
=(2c)(6c)+40cd+(5d)(5d)................................a^2+2ab+b^2=(a+b)(a+b)
=(2c+5d)(6c+5d)
=(2p)(3p)-25pq+(-7q)(-2q)................................a^2-2ab+b^2=(a-b)(a-b)
=(2p-7q)(3p-2q)
2.) 12c^2+40cd+25d^2
=(2c)(6c)+40cd+(5d)(5d)................................a^2+2ab+b^2=(a+b)(a+b)
=(2c+5d)(6c+5d)
2006-12-14 6:01 am
6p^2-25pq+14q^2
=( 3p-2q) (2p-7q)
12c^2+40cd+25d^2
= (2c-5d) (6c-5d)
2006-12-13 22:03:05 補充：
type it wrongly, the answer for Q.2 should be (2c+5d) (6c+5d)
=( 3p-2q) (2p-7q)
12c^2+40cd+25d^2
= (2c-5d) (6c-5d)
2006-12-13 22:03:05 補充：
type it wrongly, the answer for Q.2 should be (2c+5d) (6c+5d)
參考: myself
2006-12-14 5:55 am
1.) 6p^2 - 25pq + 14q^2
=(2p-7q)(3p-2q)
2.) 12c^2+40cd+25d^2
=(2c+5d)(6c+5d)
=(2p-7q)(3p-2q)
2.) 12c^2+40cd+25d^2
=(2c+5d)(6c+5d)
收錄日期: 2021-04-23 19:17:43
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https://hk.answers.yahoo.com/question/index?qid=20061213000051KK04701