標題:
一條數學題.....................thx!!!
發問:
7[2(y-3)+y] = 4[1-3(y+2)] 唔該!!!
最佳解答:
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7[2(y-3)+y] = 4[1-3(y+2)] 7(2y - 6 + y) = 4(1 - 3y - 6) 14y - 42 + 7y = 4 - 12y - 24 21y - 42 = -20 - 12y 21y + 12y = -20 + 42 33y = 22 y = 22/33 y = 2/3 希望幫到你 2006-11-26 22:49:34 補充： 上面那位朋友7[2y-6 y] = 4[1-3y 6] 2006-11-26 22:51:44 補充： 又食字...........知唔知點解成日食左 d 字.........上面那位朋友7[2y - 6y] = 4[1-3y-6]才是正確是 -6 不是 6
其他解答:
7[2(y-3)+y] = 4[1-3(y+2)] 7[2y-6+y]=4[1-3y-6] 14y-42+7y=4-12y-24 14y+7y+12y=4-24+42 33y=46-24 33y=22 2 y=--- 3|||||Take the y=??. 7[2(y-3)+y]=4[1-3(y+2)] 7(2y-6+y)=4(1-3y-6) 14y-42+7y=4-12y-24 14y+7y+12y=4-24+42 33y=22 y=22/33 The highest one have some misakes... 2006-11-27 07:44:56 補充： Add some morey=22/33Need to make it smallery=2/3|||||7[2(y-3)+y] = 4[1-3(y+2)] 7(2y-6+y) = 4(1-3y-6) 7(3y-6) = 4(-5-3y) 21y-42 = -20-12y 21y+12y = -20+42 33y = 22 y = 22/33 y = 2/3
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