Question

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

Answer 1

你可以试着模仿这2题
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

Answer 2

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

Answer 3

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