当x=3,y=-1/3时,求多项式3x²y-[2xy²-2(xy-3/2x²y)+xy]+3xy²的值
问题描述：
当x=3,y=-1/3时,求多项式3x²y-[2xy²-2(xy-3/2x²y)+xy]+3xy²的值
答
原式=3X²Y-2XY²+2XY-3X²Y-XY+3XY²
=(3-3)X²Y+(3-2)XY²+(2-1)XY
=XY²+XY
=XY(Y+1)
∵X=3 Y=-1/3
∴原式=3×(-1/3)(-1/3+1)
=-1×2/3
=-2/3