x^2-y^2-x+y = ?
麻煩STEP 詳細DD
math FACTORIZE
2008-05-04 9:53 pm
回答 (4)
2008-05-04 10:52 pm
✔ 最佳答案
x^2-y^2-x+y=(x+y)(x-y)-x+y=(x+y)(x-y)-(x-y)
=(x-y)(x+y-1)
數學有條式係: a^2-b^2=(a+b)(a-b)
2008-05-04 14:54:41 補充：
最頂o個位肯定唔啱,
你試下個數都知唔得啦
2008-05-04 14:56:25 補充：
sorry, 係啱, 不過唔係約到最簡
即係都係錯
參考: my maths book
2008-05-05 3:18 am
我甘岩做緊maths wb-3-"先睇到哩條同我做緊果d差吾多,
x^2-y^2-x y 先睇前面x^2-y^2果部份
=(x y)(x-y)-x y {因為有個恆等式 (identity)系a^2-b^2=(a b)(a-b)}
=(x y)(x-y)-1(x-y) {抽左個-1出黎,所以本身既-x y變成甘}
{點解可以抽-1?!因為-1乘咩都只會影響數值既正負號..同埋甘做就可以方便下面繼續抽common factor}
返返條數先~
=(x y)(x-y) -1(x-y) [重複廢話前既一個step]
=(x-y)[(x y)-1] {個(x-y)就甘就抽左出黎喇}
(x-y)[(x y)-1]就係最後答案喇~~
吾知你睇吾睇得明呢?!@o@
x^2-y^2-x y 先睇前面x^2-y^2果部份
=(x y)(x-y)-x y {因為有個恆等式 (identity)系a^2-b^2=(a b)(a-b)}
=(x y)(x-y)-1(x-y) {抽左個-1出黎,所以本身既-x y變成甘}
{點解可以抽-1?!因為-1乘咩都只會影響數值既正負號..同埋甘做就可以方便下面繼續抽common factor}
返返條數先~
=(x y)(x-y) -1(x-y) [重複廢話前既一個step]
=(x-y)[(x y)-1] {個(x-y)就甘就抽左出黎喇}
(x-y)[(x y)-1]就係最後答案喇~~
吾知你睇吾睇得明呢?!@o@
參考: 多謝maths牙sir專心教導
2008-05-05 2:16 am
x^2-y^2-x+y
=(x+y)(x-y) -(x-y) {because (a^2-b^2)=(a+b)(a-b)}
=(x-y)[(x+y)-1]
=(x+y)(x-y) -(x-y) {because (a^2-b^2)=(a+b)(a-b)}
=(x-y)[(x+y)-1]
參考: me
2008-05-04 10:18 pm
x^2-y^2-x+y
=(x+y)(x-y)-(x-y)
=(x-y)(x+y-1)
=(x+y)(x-y)-(x-y)
=(x-y)(x+y-1)
收錄日期: 2021-04-13 15:30:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080504000051KK01364