標題:
發問:
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!~
最佳解答:
2y + 1/3(35+y)x3 + 2/3(35-y)x4 = 90 2y + (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
其他解答:
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+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 ^_^|||||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|||||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+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|||||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
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方程!急!!!!!!!!~發問:
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!~
最佳解答:
2y + 1/3(35+y)x3 + 2/3(35-y)x4 = 90 2y + (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
其他解答:
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+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 ^_^|||||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|||||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+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|||||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
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