x+y=1,xy=-1/2,求x(x+y)(x-y)-x(x+y)2

x+y=1,xy=-1/2,求x(x+y)(x-y)-x(x+y)2
x+y=1,xy=-1/2,求x(x+y)(x-y)-x(x+y)2

x+y=1,xy=-1/2,求x(x+y)(x-y)-x(x+y)2
x(x+y)(x-y)-x(x+y)2 =x(x+y)[(x-y）－（x+y）]
=x（x＋y）（－2y）=－2xy（x＋y）
=－2×（－1/2）×1
=1

解:原式=x(x+y)(x-y-x-y)
=-2xy(x+y)
=-1

x(x+y)(x-y)-x(x+y)²
=x(x+y)(x-y-x-y)
=-2yx(x+y)
=-2×（-1/2）×1
=1a+b=2,ab=2求1/2a3b+a2b2+1/2ab31/2a³b+a²b²+1/2ab³ =1/2*ab(a²+2ab+b²) =1/2*ab(a+b)²...

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x(x+y)(x-y)-x(x+y)²
=x(x+y)(x-y-x-y)
=-2yx(x+y)
=-2×（-1/2）×1
=1

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