原式=-[xy(2x-y)+xy2]/3xy2
=-(2x-y+y)/3y
=-2x/3y不好意思，剛才少打了一個(gè)平方，原題是這樣的：(-1/3xy2)2[xy(2x-y)+xy2]，謝謝！原式=-[xy(2x-y)+xy²]/9x²y4 =-(xy.2x)/9x²y4 =-2x²y/9x²y4=-2/9y³式中除2x-y是數(shù)字2以外，其他全部是平方，可能我設(shè)置的時(shí)候沒能完全設(shè)對(duì)格式，然后第一個(gè)是三分之一=1/9x²y4[xy(2x-y+y)}=2/9x4y5
計(jì)算題：(-1/3xy2)[xy(2x-y)+xy2],
計(jì)算題：(-1/3xy2)[xy(2x-y)+xy2],
數(shù)學(xué)人氣：891 ℃時(shí)間：2020-05-11 05:47:34
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