Question

（2x²y-2xy²）-[（-3x²y²+3x²y）+（3x²y²-3xy²）]，其中x=-1，y=2．

Answer 1

原式=2x²y-2xy²-[-3x²y²+3x²y+3x²y²-3xy²]
=2x²y-2xy²+3x²y²-3x²y-3x²y²+3xy²
=2x²y-3x²y-2xy²+3xy²+3x²y²-3x²y²
=-x²y+xy²，
當x=-1，y=2時，
原式=-（-1）²×2+（-1）×2²
=-2-4
=-6．