若x

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若x若x若x(x^2+y^2)(x-y)-(x^2-y^2)(x+y)=(x^2+y^2)(x-y)(x-y)(x+y)^2=(x-y)[(x^2+y^2)-(x+y)^2]=(x-y)(x^2+y^

若x
若x

若x
(x^2+y^2)(x-y)-(x^2-y^2)(x+y)
=(x^2+y^2)(x-y)(x-y)(x+y)^2
=(x-y)[(x^2+y^2)-(x+y)^2]
=(x-y)(x^2+y^2-x^2-y^2-2xy)
=-2xy(x-y)
由x0 x-y<0
又-2<0
因此(x^2+y^2)(x-y)-(x^2-y^2)(x+y)>0
(x^2+y^2)(x-y)>(x^2-y^2)(x+y)