(2x²y-2xy²)-[(-3x²y²+3x²y)+(3x²y²-3xy²)],其中x=-1,y=2.
问题描述:
(2x²y-2xy²)-[(-3x²y²+3x²y)+(3x²y²-3xy²)],其中x=-1,y=2.
答
=2x²-2xy²+3x²y²-3x²y-3x²y²+3xy²
=-x²y+xy²
=-2-4
=-6