Maths 恆等式
2015-04-15 3:19 am
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回答 (2)
2015-04-15 3:30 am
✔ 最佳答案
(k-6-y)²= [ k-(6+y) ]² =>(a-b)²形式
= k²-2k(6+y)+(6+y)²
= k²-12k-2ky+36+12y+y²
2015-04-14 19:36:08 補充：
這題用到的公式：
(a-b)² = a²-2ab+b²
(a+b)² = a²+2ab+b²
2015-04-15 5:59 am
(-y+k-6)^2
=[(k-y)-6)^2
=(k-y)^2 -12(k-y)+36
=k^2 -2ky+y^2-12k+12y+36
=[(k-y)-6)^2
=(k-y)^2 -12(k-y)+36
=k^2 -2ky+y^2-12k+12y+36
收錄日期: 2021-04-15 19:03:45
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20150414000051KK00064