2.(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)=-(x-y)²(x-y)²-(x-y)(x+y)(x²+y²)①=-(x-y)^4-[(x²-y²)(x²+y²)]②=-（x-y)^4-[x^4-y^4]③如果对,怎么继续做,如果错了,哪里错了,第几步,请改

2.(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)=-(x-y)²(x-y)²-(x-y)(x+y)(x²+y²)①=-(x-y)^4-[(x²-y²)(x²+y²)]②=-（x-y)^4-[x^4-y^4]③如果对,怎么继续做,如果错了,哪里错了,第几步,请改
2.(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)
=-(x-y)²(x-y)²-(x-y)(x+y)(x²+y²)①
=-(x-y)^4-[(x²-y²)(x²+y²)]②
=-（x-y)^4-[x^4-y^4]③
如果对,怎么继续做,如果错了,哪里错了,第几步,请改正

2.(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)=-(x-y)²(x-y)²-(x-y)(x+y)(x²+y²)①=-(x-y)^4-[(x²-y²)(x²+y²)]②=-（x-y)^4-[x^4-y^4]③如果对,怎么继续做,如果错了,哪里错了,第几步,请改
第一步错了 (x+y)²(x-y)²=[(x+y)(x-y)]²=(x²-y²)² 而不是 -(x-y)²(x-y)²
(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)
=(x²-y²)²-(x²-y²)(x²+y²)
=(x²-y²)[(x²-y²)-(x²+y²)]
=(x²-y²)(-2y²)
=-2(x+y)(x-y)y²

2.(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)

=(x-y)(x+y）[2(x-y)(x+y）-(x²+y²)]
=(x-y)(x+y）[2(x²-y²)-(x²+y²)]
=(x-y)(x+y）(x²-3y²)

第一步错了！
=（x+y）（x-y)【（x+y）（x-y）-x^2-y^2】
祝你学习进步！！
望采纳！

未知

没看懂你第一步怎么得来的。
.(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)
=[(x+y)(x-y)[(x+y)(x-y)-(x²+y²)]
=[(x+y)(x-y)]×（-2y²）
=（x²-y²)(-2y²)
=2y^4 -2x²...

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没看懂你第一步怎么得来的。
.(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)
=[(x+y)(x-y)[(x+y)(x-y)-(x²+y²)]
=[(x+y)(x-y)]×（-2y²）
=（x²-y²)(-2y²)
=2y^4 -2x²y²

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x

要先提出一个：(x-y)(x+y)
(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)
=(x-y)(x+y)[(x-y)(x+y)-(x²+y²)]
=(x²+y²)[(x²+y²)-(x²+y²)]
=0<...

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要先提出一个：(x-y)(x+y)
(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)
=(x-y)(x+y)[(x-y)(x+y)-(x²+y²)]
=(x²+y²)[(x²+y²)-(x²+y²)]
=0
应该没错
看得懂吗？不懂可以向我提问^ ^

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第一步到第二步就错了
第二步应该是[(x+y)(x-y)]^2-(x^2-y^2)(x^2+y^2)
第三步（x^2-y^2)^2-(x^4-y^4)
第四步化开x^4-2x^2y^2+y^4-x^4+y^4
第五步就是结果了

第一步错了 (x+y)²(x-y)²=[(x+y)(x-y)]²=(x²-y²)² 而不是 -(x-y)²(x-y)²
原式=(x²-y²)²-(x-y)(x+y)(x²+y²)
=(x²-y²)[(x²-y²)...

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第一步错了 (x+y)²(x-y)²=[(x+y)(x-y)]²=(x²-y²)² 而不是 -(x-y)²(x-y)²
原式=(x²-y²)²-(x-y)(x+y)(x²+y²)
=(x²-y²)[(x²-y²)-(x²+y²)]
=(x²-y²)(-2y²)
=-2(x+y)(x-y)y²
很高兴数学团队【逻辑美】为您解答，祝您学习进步。

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.(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)
=（x+y）(x-y)[(x+y)(x-y)-(x²+y²)]
=（x²-y²）[（x²-y²）-（x²+y²）]
=（x²-y²）（-2y²）

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.(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)
=（x+y）(x-y)[(x+y)(x-y)-(x²+y²)]
=（x²-y²）[（x²-y²）-（x²+y²）]
=（x²-y²）（-2y²）
=-2y²（x²-y²）

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不对。第一步就错了。比较第一步跟原式，等于说把(x+y)²和-(x-y)²等价了。这显然是不科学的。前者是非负数，后者则是非正数。
(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)=（x-y）(x+y)[(x+y)(x-y)-(x²+y²)]
...

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不对。第一步就错了。比较第一步跟原式，等于说把(x+y)²和-(x-y)²等价了。这显然是不科学的。前者是非负数，后者则是非正数。
(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)=（x-y）(x+y)[(x+y)(x-y)-(x²+y²)]
=（x-y）(x+y)(x²-y²-x²-y²)
=（x-y）(x+y)(-2y²)
=-2y²（x-y）(x+y)

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　　第①步错了
　　(x+y)²≠-(x-y)²！

　　改正：
　　(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)
　　=[(x+y)(x-y)]²-(x-y)(x+y)(x²+y²)①
　　=[(x²-y²)]²...

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　　第①步错了
　　(x+y)²≠-(x-y)²！

　　改正：
　　(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)
　　=[(x+y)(x-y)]²-(x-y)(x+y)(x²+y²)①
　　=[(x²-y²)]²-[(x²-y²)(x²+y²)]②
　　=x^4+y^4-2x²y²-[x^4-y^4]③
　　=x^4+y^4-2x²y²-x^4+y^4
　　=2y^4-2x²y²
　　=2y²(y²-x²)
　　=2y²(x+y)(y-x).

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.(x+y)²(x-y)²-(x-y)(x+y）(x²+y²)
原式=(x+y)(x-y)×[(x+y)(x-y)-(x²+y²）]【把(x+y)(x-y）提公因式】
=（(x²-y²)×(x²-y²-x²-y²)
=x²y²-y^4