将x(x+y)(x-y)-x(x+y)²进行因式分解并求当x+y=1,xy=-2分之1时原式的值
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将x(x+y)(x-y)-x(x+y)²进行因式分解并求当x+y=1,xy=-2分之1时原式的值
将x(x+y)(x-y)-x(x+y)²进行因式分解并求当x+y=1,xy=-2分之1时原式的值
将x(x+y)(x-y)-x(x+y)²进行因式分解并求当x+y=1,xy=-2分之1时原式的值
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将x(x+y)(x-y)-x(x+y)2进行因式分解,并求当x+y=1007,xy=-1时此式的值
解
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)×1007
=2×1007
=2014
利用因式分解求x(x+y)(x-y)-x(x+y)^2的值,其中x+y=1,xy=1/2.
解∶x﹙x+y﹚﹙x-y﹚-x﹙x+y﹚²
=x﹙x+y﹚﹙x-y﹚-x﹙x+y﹚﹙x+y﹚
=x﹙x+y﹚[﹙x-y﹚-﹙x+y﹚]
=x﹙x+y﹚﹙x-y-x-y﹚
=x﹙x+y﹚﹙﹣2y﹚
=﹙﹣2xy﹚﹙x+y﹚
=﹙﹣2﹚× ½× 1
=﹣1
解
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
x(x+y)(x-y)-x(x+y)²
=x(x+y)[(x-y)-(x+y)]
=-2xy(x+y)
当x+y=1,xy=-2分之1时
原式=-2×(-2分之1)×1=1
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