我想問可5可以比這條式的步式我:
2y+1/3(35+y)×3+2/3(35-y)×4=90
我因為果日冇番學，所以抄不到老師比的步式和答案!~
方程!急!!!!!!!!~
2006-10-29 6:33 am
回答 (7)
2006-10-29 6:46 am
✔ 最佳答案
2y + 1/3(35+y)x3 + 2/3(35-y)x4 = 902y + (35+y) + 8/3(35-y) = 90
6y + 105 + 3y + 8(35-y) = 270
6y + 105 + 3y + 190 - 8y = 270
6y + 3y - 8y = 270 -105 - 190
y = -25
參考: Hehe... me lor
2006-10-29 7:28 am
2y+1/3(35+y)*3+2/3(35-y)*4
2y+(35/3+y/3)*3+(70/3-2y/3)*4=90
2y+35+y+280/3-8y/3=90
3y-8y/3=90-35-280/3
(3y-8y/3)*3=(90-35-280/3)*3
9y-8y=270-105-280
y=負115
2y+(35/3+y/3)*3+(70/3-2y/3)*4=90
2y+35+y+280/3-8y/3=90
3y-8y/3=90-35-280/3
(3y-8y/3)*3=(90-35-280/3)*3
9y-8y=270-105-280
y=負115
2006-10-29 7:04 am
2y+1/3(35+y)×3+2/3(35-y)×4=90
可以分開嚟解
first : 1/3(35+y)x3=35+y
second : 2/3(35-y)x4=8/3(35-y)=(280/3)-(8/3)y
third : 把他們全部相加 得到 2y+y-(8/3)y=90-35-(280/3)
four : (1/3)y=-(115/3)
so y= -115
^_^
可以分開嚟解
first : 1/3(35+y)x3=35+y
second : 2/3(35-y)x4=8/3(35-y)=(280/3)-(8/3)y
third : 把他們全部相加 得到 2y+y-(8/3)y=90-35-(280/3)
four : (1/3)y=-(115/3)
so y= -115
^_^
2006-10-29 6:50 am
2y+1/3(35+y)*3+2/3(35-y)*4=90
{所有數乘到分母 = 3 (所有數分母既L.C.M.) 黎通分母}
(你去通分母and消失左 d 分母之後, 條數就會易計好多架喇).
Therefore, 2y * 3 + (1/3)(35+y) * 3 * 3 + (2/3)(35-y) * 4 * 3 = 90 * 3
6y + 105 + 3y + 280 - 8y = 270
y + 385 = 270
y = - 115
{所有數乘到分母 = 3 (所有數分母既L.C.M.) 黎通分母}
(你去通分母and消失左 d 分母之後, 條數就會易計好多架喇).
Therefore, 2y * 3 + (1/3)(35+y) * 3 * 3 + (2/3)(35-y) * 4 * 3 = 90 * 3
6y + 105 + 3y + 280 - 8y = 270
y + 385 = 270
y = - 115
2006-10-29 6:48 am
2y+1/3(35+y)×3+2/3(35-y)×4=90
2y+35+y+280/3-8y/3=90
3y+280/3-8y/3=55
(9y+280-8y)/3=55
y/3+280/3=55
y/3=165/3-280/3
y/3=-115/3
y=-115
提一提你！！！
上面第一的式子：：：：：
6y + 105 + 3y + 280 - 8y = 270
y + 385 = 270
y=-105有問題
385-270是-115而不是 -105
2y+35+y+280/3-8y/3=90
3y+280/3-8y/3=55
(9y+280-8y)/3=55
y/3+280/3=55
y/3=165/3-280/3
y/3=-115/3
y=-115
提一提你！！！
上面第一的式子：：：：：
6y + 105 + 3y + 280 - 8y = 270
y + 385 = 270
y=-105有問題
385-270是-115而不是 -105
2006-10-29 6:38 am
2y+1/3(35+y)×3+2/3(35-y)×4=90
3[2y+1/3(35+y)x3+2/3(35-y)x4=90x3
6y+3(35+y)+8(35-y)=270
6y+105+3y+280-8y=270
6y+3y-8y=270-105-280
y=270-380
y=-110
2006-10-28 22:40:12 補充：
y=270-380y=-110 改做y=270-385y=-105
3[2y+1/3(35+y)x3+2/3(35-y)x4=90x3
6y+3(35+y)+8(35-y)=270
6y+105+3y+280-8y=270
6y+3y-8y=270-105-280
y=270-380
y=-110
2006-10-28 22:40:12 補充：
y=270-380y=-110 改做y=270-385y=-105
2006-10-29 6:36 am
2y+1/3(35+y)×3+2/3(35-y)×4=90
6y +3 (35 + y) + 8 (35 - y) = 270
6y + 105 + 3y + 280 - 8y = 270
y + 385 = 270
y = - 105
6y +3 (35 + y) + 8 (35 - y) = 270
6y + 105 + 3y + 280 - 8y = 270
y + 385 = 270
y = - 105
收錄日期: 2021-04-12 18:09:42
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061028000051KK05464