4(y-2z)(2z+3y)
=4(y)+(4)(-2z)+4(2z)+4(3y)
=4y-8z+8z+12y
=16y2-16z2
Is the step right??
If NOT,please tell me...
Maths question
2006-10-26 2:09 am
回答 (5)
2006-10-26 2:19 am
✔ 最佳答案
4(y-2z)(2z+3y)=4(y)+(4)(-2z)+4(2z)+4(3y)
第一步已經錯了
括號要逐個拆
應用如下的步驟：
4(y-2z)(2z+3y)
= [4(y) – 4(2z)] (2z+3y)
= (4y – 8z)(2z + 3y)
= 4y(2z + 3y) – 8z(2z + 3y)
= 4y(2z) + (4y)(3y) – 8z(2z) – 8z(3y)
= 8yz + 12y^2 – 16z^2 – 24zy
= 12y^2 – 16z^2 – 16zy
2006-10-26 11:36 am
4(y-2z)(2z+3y)=(4y-8z)(2z+3y)
= (8yz+12y^2-16z^2-24yz)
= 12y^2-16yz-16z^2
= 4(3y^2-4yz-4z^2)
=4(3y+2z)(y-2z)
= (8yz+12y^2-16z^2-24yz)
= 12y^2-16yz-16z^2
= 4(3y^2-4yz-4z^2)
=4(3y+2z)(y-2z)
2006-10-26 2:15 am
你想計咩?
如果要最簡既就係抽個16出黎
16(y2-z2)
如果要最簡既就係抽個16出黎
16(y2-z2)
2006-10-26 2:14 am
It has some wrong........
4(y-2z)(2z+3y)
=(4y-8y)(2z+3y)
=8zy+12y^2-16zy-24y^2
=-12Y^2-8ZY
4(y-2z)(2z+3y)
=(4y-8y)(2z+3y)
=8zy+12y^2-16zy-24y^2
=-12Y^2-8ZY
2006-10-26 2:14 am
錯左la~第1步就錯左la~
4(y-2z)(2z+3y)
=(4y-8y)(2z+3y)
=8zy+12y^2-16zy-24y^2
=-12Y^2-8ZY
4(y-2z)(2z+3y)
=(4y-8y)(2z+3y)
=8zy+12y^2-16zy-24y^2
=-12Y^2-8ZY
收錄日期: 2021-04-12 18:13:33
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061025000051KK02633