已知x+y=1,xy=-二分之一,求x(x+y)(x-y)-x(x+y)²

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已知x+y=1,xy=-二分之一,求x(x+y)(x-y)-x(x+y)²

原式=x(x+u)(x-y-x-y)
=-2xy(x+y)
=-2×(-1/2)×1
=1

原式=(x+y)(x2-xy-x2-xy)=(x+y)(-2xy)=1