the answer of this 3 questions include steps:
1) 3(a-b)(3r-t)+9(b-a)(2t-5s)
2) r(c-d)二次方-s(c-d)二次方
3) 4(a-b)三次方-12(b-a)二次方
F.2 mathz
2007-10-12 10:15 pm
回答 (2)
2007-10-12 10:33 pm
✔ 最佳答案
1)3(a-b)(3r-t) + 9(b-a)(2t-5s)
= 3(a-b)(3s-t) + 9(-a+b)(2t-5s) <-- ( b-a -> -a+b )
= 3(a-b)(3s-t) - 9(a-b)(2t-5s)
= 3(a-b)[ 3s-t - 3(2t-5s) ] <-- Common factor = 3(a-b)
= 3(a-b)(3s - t - 6t + 15s)
= 3(a-b)(18s-7t)
2)
r(c-d)^2 - s(c-d)^2
= (c-d)^2(r-s) <-- Common factor = (c-d)^2
3)
4(a-b)^3 - 12(b-a)^2
= 4(a-b)^3 - 12(-a+b)^2
= 4(a-b)^3 - 12(-1)^2(a-b)^2 <-- (-a+b)^2 = [ (-1)(a-b) ]^2 = (-1)^2(a-b)^2
= 4(a-b)^3 - 12(a-b)^2
= 4(a-b)^2 [ (a-b) - 3 ]
= 4(a-b)^2 (a-b-3)
參考: Me
2007-10-12 10:45 pm
1) 3(a-b)(3r-t) + 9(b-a)(2t-5s) = {3(a-b)}{(3r-t)-3(2t-5s)
=3(a-b)(3r+5s-7t)
2) r(c-d)^2 - s(c-d)^2 = (c-d)^2 (r-s)
3) 4(a-b)^3 - 12(b-a)^2 = 4(a-b)^2{(a-b)-3} [Note (a-b)^2=(b-a)^2]
=4(a-b)^2(a-b-3)
=3(a-b)(3r+5s-7t)
2) r(c-d)^2 - s(c-d)^2 = (c-d)^2 (r-s)
3) 4(a-b)^3 - 12(b-a)^2 = 4(a-b)^2{(a-b)-3} [Note (a-b)^2=(b-a)^2]
=4(a-b)^2(a-b-3)
收錄日期: 2021-04-13 19:03:39
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20071012000051KK01369