(负三分之一 xy)²[xy(2x-y)-2x(xy-y²)]
=9分之1*x²y²*[xy(2x-y) - 2xy(x-y)]
=9分之1*x²y²*[xy(2x-y - 2x+2y)]
=9分之1*(xy)³y
已知x=-一又二分之一,y=-2,那么:xy=2分之3×2=3
所以:原式=9分之1*27*(-2)=-6
(负三分之一 xy)²[xy(2x-y)-2x(xy-y²)]
=9分之1*x²y²*[xy(2x-y) - 2xy(x-y)]
=9分之1*x²y²*[xy(2x-y - 2x+2y)]
=9分之1*(xy)³y
已知x=-一又二分之一,y=-2,那么:xy=2分之3×2=3
所以:原式=9分之1*27*(-2)=-6