问题标题：
[2x²-(x+y)(x-y)][(-x-y)(y-x)+2y²]其中x=1,y=2数学：[2x²-(x+y)(x-y)][(-x-y)(y-x)+2y²]其中x=1，y=2
问题描述：
[2x²-(x+y)(x-y)][(-x-y)(y-x)+2y²]其中x=1,y=2
数学：[2x²-(x+y)(x-y)][(-x-y)(y-x)+2y²]其中x=1，y=2
华顺刚回答：
[2x²-(x+y)(x-y)][(-x-y)(y-x)+2y²]=[2x²-(x²-y²)][(x+y)(x-y)+2y²]=(2x²-x²+y²)(x²-y²+2y²)=(x²+y²)²=(1²+2²)²=5&...
裴燕玲回答：
请问一下就是原式那个答案在哪呢？25是正确答案，25的上一个式子的答案在哪里？
华顺刚回答：
[2x²-(x+y)(x-y)][(-x-y)(y-x)+2y²]=[2x²-(x²-y²)][(x+y)(x-y)+2y²]=(2x²-x²+y²)(x²-y²+2y²)=(x²+y²)²=(1²+2²)²=(1+4)²=5²=25
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