(2x²y-2xy²)-[(-3x²y²+3x²y)+(3x²y²-3xy²)]其中x=-1y=2先化简,再求值
问题描述:
(2x²y-2xy²)-[(-3x²y²+3x²y)+(3x²y²-3xy²)]其中x=-1y=2
先化简,再求值
答
xy2-x2y =16
答
(2x²y-2xy²)-[(-3x²y²+3x²y)+(3x²y²-3xy²)]
=2x^2y-2xy^2-(3x^2y-3xy^2)
=xy^2-x^2y
=xy(y-x)
=-1*2(2+1)
=-6
答
原式=2x²y-2xy²-(-3x²y²+3x²y+3x²y²-3xy²)
=2x²y-2xy²-(3x²y-3xy²)
=2x²y-2xy²-3x²y+3xy²
=-x²y+xy²
=-2-4
=-6