(a+b)^2-a-b
更新1:
點解要抽個1既?
中三因式分解 快 ２0分
回答 (4)
2010-01-09 1:26 am
✔ 最佳答案
(a+b)^2-a-b=(a+b)^2-(a+b)
=(a+b)[(a+b)-1]
=(a+b)(a+b-1)
為何要1呢?
這是因為它有(a+b)這個因子。
而1x(a+b)=a+b,
所以要x1。
你可以把答案Expand開去,會得出相同答案。
希望幫到你!!!!
參考: 自己
2010-01-10 1:33 am
the factor(s) of (a+b)^2 : (a+b)(a+b)
the factor(s) of -a-b=-(a+b) :(-1)*(a+b)
both of them have a common factor (a+b),
so we need to combine them.
on the other hand,we need to add their remaining factor(s) [(a+b) and (-1)]
so the result of the factorization is: (a+b)[(a+ b)+(-1)]
=(a+b)(a+b-1)
the factor(s) of -a-b=-(a+b) :(-1)*(a+b)
both of them have a common factor (a+b),
so we need to combine them.
on the other hand,we need to add their remaining factor(s) [(a+b) and (-1)]
so the result of the factorization is: (a+b)[(a+ b)+(-1)]
=(a+b)(a+b-1)
參考: my mind
2010-01-08 7:22 pm
(a+b)^2-a-b
= (a + b) (a + b) - a - b
= (a + b) (a + b) - (a + b)
= (a + b) [(a + b) - 1]
= (a + b) (a + b - 1)
2010-01-08 11:23:55 補充：
點解要抽個1既?
Because (a + b) = (a + b) X 1
= (a + b) (a + b) - a - b
= (a + b) (a + b) - (a + b)
= (a + b) [(a + b) - 1]
= (a + b) (a + b - 1)
2010-01-08 11:23:55 補充：
點解要抽個1既?
Because (a + b) = (a + b) X 1
參考: me
2010-01-08 7:06 am
(a + b)^2 - a - b
= (a + b)^2 - (a + b)
= (a + b)^2 - (a + b)(1)
= (a + b)(a + b - 1)
2010-01-07 23:15:06 補充：
Let a+b = x :
x^2 - x
= x^2 - 1x....1x可不寫,清楚些罷了
= x(x - 1)
= (a + b)^2 - (a + b)
= (a + b)^2 - (a + b)(1)
= (a + b)(a + b - 1)
2010-01-07 23:15:06 補充：
Let a+b = x :
x^2 - x
= x^2 - 1x....1x可不寫,清楚些罷了
= x(x - 1)
收錄日期: 2021-04-21 22:13:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20100107000051KK01709