Question

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

先化简,再求值

Answer 1

原式=2x²y-2xy²-(-3x²y²+3x²y+3x²y²-3xy²)
=2x²y-2xy²-(3x²y-3xy²)

=2x²y-2xy²-3x²y+3xy²
=-x²y+xy²
=-2-4
=-6

Answer 2

(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

Answer 3

xy2-x2y =16