已知x+y=8,xy=12,求(1)x2y+xy2(2)x2-xy+y2(3)x-y

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(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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