-2(2x²y-2xy²)-[(-3x²y²+3x²y)+(3
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-2(2x²y-2xy²)-[(-3x²y²+3x²y)+(3²y²-2xy²)],其中x=-1,y=2
因为x=-1,y=2,所以:
-2(2x²y-2xy²)-[(-3x²y²+3x²y)+(3x²y²-2xy²)]
=-4x²y+4xy²-(3x²y-2xy²)
=-4x²y+4xy²-3x²y+2xy²
=-7x²y+6xy²
=xy(-7x+6y)
=-1*2*(7+12)
=-38
-2(2x²y-2xy²)-[(-3x²y²+3x²y)+(3x²y²-2xy²)]
=-4x²y+4xy²-(3x²y-2xy²)
=-4x²y+4xy²-3x²y+2xy²
=-7x²y+6xy²
=xy(-7x+6y)
=-1*2*(7+12)
=-38
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