4xy-((x^2+5xy-y^2)-(x^2+3xy-2y^2)),其中x=2,y=-1.

biaozhunbaoxiu2022-10-04 11:39:541条回答

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yueliu622 共回答了19个问题 | 采纳率89.5%
4xy-((x^2+5xy-y^2)-(x^2+3xy-2y^2))
=4xy-(x²+5xy-y²)+(x²+3xy-2y²)
=4xy-x²-5xy+y²+x²+3xy-2y²
=2xy-y²
当x=2,y=-1时
原式=-4-1=-5
1年前

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先化简再求值:4xy-[(x^2+5xy-y^2)-(x^2+3xy-2y^2)]其中x=1/4,y=1/2
开的1年前1
普通gg 共回答了21个问题 | 采纳率81%
4xy-[(x^2+5xy-y^2)-(x^2+3xy-2y^2)]
=4xy-(x^2+5xy-y^2)+(x^2+3xy-2y^2)
=4xy-x^2-5xy+y^2+x^2+3xy-2y^2
=2xy-y^2
当x=1/4,y=1/2时
原式=1/4-1/4
=0
4xy-{(x^2+5xy-y^2)-(x^2+3xy-2y^2)}其中x=-1/4y=-1/2
冰雕小猫1年前1
kokom 共回答了13个问题 | 采纳率84.6%
因为x=-1/4,y=-1/2,所以:
4xy-{(x²+5xy-y²)-(x²+3xy-2y²)}
=4xy-(x²+5xy-y²-x²-3xy+2y²)
=4xy-(2xy+y²)
=2xy-y²
=y(2x-y)
=(-1/2)*[2*(-1/4)-(-1/2)]
=(-1/2)*0
=0