中一1條方程..
回答 (5)
2011-02-08 3:45 am
✔ 最佳答案
2[3y-(4-2y)]=4-2y2[3y-4+2y]=4-2y
6y-8+4y=4-2y
6y+4y+2y=4+8
12y=12
y=1
2011-02-08 5:43 am
2[3y-(4-2y)]=4-2y
3y-(4-2y)=2-y
3y-4+2y=2-y
5y-4=2-y
6y-4=2
6y=6
y=1
3y-(4-2y)=2-y
3y-4+2y=2-y
5y-4=2-y
6y-4=2
6y=6
y=1
2011-02-08 3:45 am
2[3y-(4-2y)]=4-2y
6y-2(4-2y)=4-2y
6y-8+4y=4-2y
12y=12
y=1
6y-2(4-2y)=4-2y
6y-8+4y=4-2y
12y=12
y=1
2011-02-08 3:36 am
2[3y-(4-2y)]=4-2y
2(3y-4+2y)=4-2y
6y-8+4y=4-2y
6y+4y-2y=4+8
8y=12
y=12/8
y=1.5
hope I can help you
2(3y-4+2y)=4-2y
6y-8+4y=4-2y
6y+4y-2y=4+8
8y=12
y=12/8
y=1.5
hope I can help you
參考: me
2011-02-08 3:34 am
2[3y-(4-2y)]=4-2y
2[3y-4+2y] =4-2y
2[5y-4]=4-2y
10y-8=4-2y
12y=12
y=1
2[3y-4+2y] =4-2y
2[5y-4]=4-2y
10y-8=4-2y
12y=12
y=1
參考: myself
收錄日期: 2021-04-16 12:14:48
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20110207000051KK01172