F.2 Maths
回答 (2)
2013-08-19 4:35 am
✔ 最佳答案
1. (x - 2y)^2 + (x + 2y)^2= x^2 - 4xy + 4y^2 + x^2 + 4xy + 4y^2
= 2x^2 + 8y^2
2(x - 2y)(x + 2y) + 16y^2
= 2(x^2 + 2xy - 2xy - 4y^2) + 16y^2
= 2(x^2 - 4y^2) + 16y^2
= 2x^2 - 8y^2 + 16y^2
= 2x^2 + 8y^2
(x - 2y)^2 + (x + 2y)^2 = 2(x - 2y)(x + 2y) + 16y^2
is identity .
2013-08-19 4:35 am
(x－2y)^2+(x+2y)^2=2(x－2y)(x+2y)+16y^2 (is it identity?)
Sol
左=(x－2y)^2+(x+2y)^2
=(x^2－4xy+4y^2)+(x^2+4xy+4y^2)
=2x^2+8y^2
右=2(x－2y)(x+2y)+16y^2
=2(x^2－4y^2)+16y^2
=2x^2+8y^2
左=右
Yes
Sol
左=(x－2y)^2+(x+2y)^2
=(x^2－4xy+4y^2)+(x^2+4xy+4y^2)
=2x^2+8y^2
右=2(x－2y)(x+2y)+16y^2
=2(x^2－4y^2)+16y^2
=2x^2+8y^2
左=右
Yes
收錄日期: 2021-04-30 17:56:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20130818000051KK00264