1/2(y+1)+1/3(y+2)+1/4(y+3)-3=0

题目

1/2(y+1)+1/3(y+2)+1/4(y+3)-3=0
不需要解了,我看了答案,我只想大家帮我分析一下
[1/2(y+1)-1]+[1/3(y+2)-1]+[1/4(y+3)-3]=0
即1/2(y-1)+1/3(y-1)+1/4(y-1)=0 ——就是这步,
我已经说了，答案都写出来了！还需要你们化简吗，就是告诉我那一步为什么这样写，不过也谢谢你

答案

你的题目这么写容易造成别人的误1/2(y+1)写成(y+1)/2更确切.
1/2(y+1)+1/3(y+2)+1/4(y+3)-3
=(y+1)/2+(y+2)/3+(y+3)/4-3
=[(y+1)/2-1+(y+2)/3-1+(y+3)/4-1]
=[(y+1)/2-(2/2)+(y+2)/3-(3/3)+(y+3)/4-(4/4)]
=(y+1-2)/2+(y+2-3)/3+(y+3-4)/4
=(y-1)/2+(y-1)/3+(y-1)/4
其实你看不懂的那步,简单的说就是把3拆成3个1,然后分别通分计算