Question

求下列整式的值(xy-3/2y-1/2)-(xy-3/2x+1)过程要完整。懂。

Answer 1

原式=xy-3/2y-1/2-xy+3/2x-1
=(xy-xy)+3/2x-3/2y-(1+1/2)
=3/2x-3/2y-3/2

追问

奇怪的说不是这样嘛。
(xy-3/2y-1/2)-(xy-3/2x+1)

=xy-3/2y-1/2-xy+3/2x-1
=(xy-xy)+3/2y+3/2x+(-1/2-1)
=3/2y+3/2x-3/2
我哪里错了帮帮忙。

追答

你好：同学！你的递等式中第二步出错了，你写的是3/2y，应该有负号

追问

那应该是-3/2y+3/2x-3/2啊

追答

加法有交换律，-3/2y+3/2x-3/2=3/2x-3/2y-3/2

追问

哈哈，我小学数学没学好，现在初二正在努力呢。

追答

你好：同学！刚才我最后也出错了，应该是-3/2y+3/2x-3/2=3/2x+3/2y-3/2

Answer 2

?êìa213?￡o?óò?é???ê?μ??μ (xy-3/2y-1/2)-(xy-3/2x+1) =xy-3/2y-1/2-xy+3/2x-1 =-3/2y-1/2+3/2x-1 =3/2(x-y)-3/2 =3

Answer 3

楼主正确