已知xy=-2,x-y=3求(x+y)(x-y)-y^2-(6x^2y-2xy^2)/2y的值。 5
3个回答
2012-03-04
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(x+y)(x-y)-y²-(6x²y-2xy²)/2y
=x²-y²-y²-3x²+xy
=-2y²-2x²+xy
=-2(x²+y²)+xy
由于xy=-2,x-y=3,所以x²+y²=(x-y)²+2xy=9-4=5
将xy=-2,x²+y²=5代入上式得到:
(x+y)(x-y)-y²-(6x²y-2xy²)/2y
=-2(x²+y²)+xy
=-2*5-2
=-12
=x²-y²-y²-3x²+xy
=-2y²-2x²+xy
=-2(x²+y²)+xy
由于xy=-2,x-y=3,所以x²+y²=(x-y)²+2xy=9-4=5
将xy=-2,x²+y²=5代入上式得到:
(x+y)(x-y)-y²-(6x²y-2xy²)/2y
=-2(x²+y²)+xy
=-2*5-2
=-12
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由xy=-2,x-y=3可得:x=1,y=-2或x=2,y=-1
当x=1,y=-2时,(x+y)(x-y)-y^2-(6x^2y-2xy^2)/2y=-1*3-4-(-12-8)/(-4)=-6
当x=2,y=-1时,(x+y)(x-y)-y^2-(6x^2y-2xy^2)/2y=3-1-(-24-4)/(-2)=-12
当x=1,y=-2时,(x+y)(x-y)-y^2-(6x^2y-2xy^2)/2y=-1*3-4-(-12-8)/(-4)=-6
当x=2,y=-1时,(x+y)(x-y)-y^2-(6x^2y-2xy^2)/2y=3-1-(-24-4)/(-2)=-12
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因为xy=-2,x-y=3
所以x=1,y=-2或x=2,y=-1(代入法)
当x=1,y=-2时,(x+y)(x-y)-y^2-(6x^2y-2xy^2)/2y=-1*3-4-(-12-8)/(-4)=-6
当x=2,y=-1时,(x+y)(x-y)-y^2-(6x^2y-2xy^2)/2y=3-1-(-24-4)/(-2)=-12
所以x=1,y=-2或x=2,y=-1(代入法)
当x=1,y=-2时,(x+y)(x-y)-y^2-(6x^2y-2xy^2)/2y=-1*3-4-(-12-8)/(-4)=-6
当x=2,y=-1时,(x+y)(x-y)-y^2-(6x^2y-2xy^2)/2y=3-1-(-24-4)/(-2)=-12
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