已知x+y=8,xy=12,求(1)x2y+xy2(2)x2-xy+y2(3)x-y
2个回答
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解：
(1)
x^2y+xy^2
=xy(x+y)
=12*8
=96
(2)
x^2-xy+y^2
=x^2+2xy+y^2-3xy
=(x+y)^2-3xy
=8^2-3*12
=64-36
=28
(3)
(x-y)^2
=x^2-2xy+y^2
=x^2+2xy+y^2-4xy
=(x+y)^2-4xy
=8^2-4*12
=64-48
=16
x-y=±4
(1)
x^2y+xy^2
=xy(x+y)
=12*8
=96
(2)
x^2-xy+y^2
=x^2+2xy+y^2-3xy
=(x+y)^2-3xy
=8^2-3*12
=64-36
=28
(3)
(x-y)^2
=x^2-2xy+y^2
=x^2+2xy+y^2-4xy
=(x+y)^2-4xy
=8^2-4*12
=64-48
=16
x-y=±4
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解析-
(1)x²y+xy²=xy(x+y)
=12x8
=96
(2)x²-xy+y²=(x+y)²-3xy
=64-3x12
=64-36
=28
(3)x-y=±√（x-y)²=√(x+y)²-4xy
=±√64-4x12
=±√16
=±4
(1)x²y+xy²=xy(x+y)
=12x8
=96
(2)x²-xy+y²=(x+y)²-3xy
=64-3x12
=64-36
=28
(3)x-y=±√（x-y)²=√(x+y)²-4xy
=±√64-4x12
=±√16
=±4
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