2(x+y-1)^2-4(1-x-y),,factorize,,有冇得計呀??
如果有,,how to do it ???
2(x+y-1)^2-4(1-x-y),,factorize,,有冇得計呀??
2006-11-06 5:52 pm
回答 (4)
2006-11-06 5:55 pm
✔ 最佳答案
2(x+y-1)^2-4(1-x-y)＝2(x+y-1)^2+4(x+y-1)
＝2(x+y-1)(x+y-1+2)
＝2(x+y-1)(x+y+1)
參考: 就咁計出來，留意只係正負號變位而已
2006-11-08 7:20 am
呢題要用到您的其中一個問題的技巧:(5x-3)(x-y)-3x(y-x),,factorize,,plx!
-x(y-x)
=x(x-y)
呢方面唔多解釋了~
2(x+y-1)^2-4(1-x-y)
=2(x+y-1)^2+4(-1+x+y)<---出面個正負號轉左,括號內要轉正負號
=2(x+y-1)^2+4(x+y-1)
=2(x+y-1)(x+y-1+4)<---抽(x+y-1),個<2>唔使理
=2(x+y-1)(x+y-3)
-x(y-x)
=x(x-y)
呢方面唔多解釋了~
2(x+y-1)^2-4(1-x-y)
=2(x+y-1)^2+4(-1+x+y)<---出面個正負號轉左,括號內要轉正負號
=2(x+y-1)^2+4(x+y-1)
=2(x+y-1)(x+y-1+4)<---抽(x+y-1),個<2>唔使理
=2(x+y-1)(x+y-3)
2006-11-06 6:01 pm
2(x+y-1)^2-4(1-x-y)
= 2(x+y-1)^2+4(x+y-1)
= (x+y-1) [2(x+y-1)+4]
= 2(x+y-1) [(x+y-1)+2]
= 2(x+y-1) (x+y-1+2)
= 2(x+y-1) (x+y+1)
= 2(x+y-1)^2+4(x+y-1)
= (x+y-1) [2(x+y-1)+4]
= 2(x+y-1) [(x+y-1)+2]
= 2(x+y-1) (x+y-1+2)
= 2(x+y-1) (x+y+1)
參考: I think this is simpliest ga la
2006-11-06 5:55 pm
2(x+y-1)^2-4(1-x-y)
=2(x+y)^2-4(x+y)+2-4+4(x+y)
=2(x+y)^2-2
=2((x+y)^2-1)
=2(x+y+1)(x+y-1)
2006-11-06 10:01:26 補充：
提示：1. 把(x+y) 當成一個 term 咁去計就會易好多！2. 使用負負得正3. 應用到 a^2-b^2 = (a+b)(a-b)
=2(x+y)^2-4(x+y)+2-4+4(x+y)
=2(x+y)^2-2
=2((x+y)^2-1)
=2(x+y+1)(x+y-1)
2006-11-06 10:01:26 補充：
提示：1. 把(x+y) 當成一個 term 咁去計就會易好多！2. 使用負負得正3. 應用到 a^2-b^2 = (a+b)(a-b)
收錄日期: 2021-04-12 22:28:23
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061106000051KK00728