题目内容
(1+y)2-2x2(1+y2)+x4(1-y)2.
考点:因式分解
专题:
分析:首先补项2x2(1-y2)进而利用完全平方公式以及平方差公式分解因式进而得出答案.
解答:解:原式=(1+y)2+2x2(1-y2)+x4(1-y2)-2x2(1-y2)-2x2(1+y2)
=[1+y+x2(1-y)]2-2x2(1-y2+1+y2)
=(x2-x2y+y+1)2-4x2
=(x2-x2y+y+1+2x)(x2-x2y+y+1-2x)
=[(x2+2x+1)-y(x2-1)][(x2-2x+1)-y(x2-1)]
=[(x+1)2-y(x2-1)][(x-1)2-y(x2-1)]
=(x+1)(x+1-xy+y)(x-1)(x-1-xy-y).
=[1+y+x2(1-y)]2-2x2(1-y2+1+y2)
=(x2-x2y+y+1)2-4x2
=(x2-x2y+y+1+2x)(x2-x2y+y+1-2x)
=[(x2+2x+1)-y(x2-1)][(x2-2x+1)-y(x2-1)]
=[(x+1)2-y(x2-1)][(x-1)2-y(x2-1)]
=(x+1)(x+1-xy+y)(x-1)(x-1-xy-y).
点评:此题主要考查了因式分解法的应用,熟练利用完全平方公式分解因式是解题关键.
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