2个回答
展开全部
解原式
=3xy²-2(xy-3/2x²y)+(3x²y-2xy²)
=3xy²-2xy+3x²y+3x²y-2xy²
=xy²+6x²y-2xy
=xy(y+6x-2)
=(-2)×(1/2)[1/2+6×(-2)-2]
=-(1/2-14)
=14-1/2
=27/2
=3xy²-2(xy-3/2x²y)+(3x²y-2xy²)
=3xy²-2xy+3x²y+3x²y-2xy²
=xy²+6x²y-2xy
=xy(y+6x-2)
=(-2)×(1/2)[1/2+6×(-2)-2]
=-(1/2-14)
=14-1/2
=27/2
追问
还有一题,已知2x-3y=1,试求(2x²+4x-3y)+2(-x²+x-3y)-17的值。
追答
解由(2x²+4x-3y)+2(-x²+x-3y)-17
=(2x²+4x-3y)+(-2x²+2x-6y)-17
=6x-9y-17
=3(2x-3y)-17
=3*1-17
=-14.
展开全部
=3xy²-2xy+3x²y+3x²y-2xy²
=(3xy²-2xy²)-2xy+(3x²y+3x²y)
=xy²-2xy+6x²y
=xy(y-2+6x)
=-2×1/2(1/2-2-12)
=-(1/2-14)
=13又2分之1
=(3xy²-2xy²)-2xy+(3x²y+3x²y)
=xy²-2xy+6x²y
=xy(y-2+6x)
=-2×1/2(1/2-2-12)
=-(1/2-14)
=13又2分之1
追问
还有一题,已知2x-3y=1,试求(2x²+4x-3y)+2(-x²+x-3y)-17的值。
追答
=(2x²-2x²)+(4x+2x)+(-3y-6y)-17
=6x-9y-17
=3(2x-3y)-17
=3×1-17
=-14
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