(x+y+z)(x-y+z)-(y-x+z)(y-x-z)
3个回答
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(x+y+z)(x-y+z)-(y-x+z)(y-x-z)
=(x+z)^2 - y^2 - (y-x)^2 + z^2
=(x+z)^2 - (y-x)^2 + z^2 - y^2
= (x+z-y+x)(x+z+y-x) + (z-y)(z+y)
=(2x +z -y)(y+z) + (z-y)(z+y)
=(z+y)(2x+z-y+z-y)
=2(z+y)(x-y+z)
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=(x+z)^2 - y^2 - (y-x)^2 + z^2
=(x+z)^2 - (y-x)^2 + z^2 - y^2
= (x+z-y+x)(x+z+y-x) + (z-y)(z+y)
=(2x +z -y)(y+z) + (z-y)(z+y)
=(z+y)(2x+z-y+z-y)
=2(z+y)(x-y+z)
【秋风燕燕为您答题 O(∩_∩)O,肯定对 】
有什么不明白可以对该题继续追问,随时在线等
如果我的回答对你有帮助,请及时选为满意答案,谢谢
追问
可以用提公因式法吗
追答
(x+y+z)(x-y+z)-(y-x+z)(y-x-z)
=(x+y+z)(x-y+z)+(y-x+z)(x-y+z)
=(x-y+z)(x+y+z+y-x+z)
=(x-y+z)(2y+2z)
=2(x-y+z)(y+z)
展开全部
解:原式=(x+y+z)(x-y+z)+(y-x+z)(x-y+z)
=(x-y+z)(x+y+z+y-x+z)
=(x-y+z)(2y+2z)
=2(x-y+z)(y+z)
=(x-y+z)(x+y+z+y-x+z)
=(x-y+z)(2y+2z)
=2(x-y+z)(y+z)
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展开全部
(x+y+z)(x-y+z)-(y-x+z)(y-x-z)
=(x+y+z)(x-y+z)-(x-y-z)(x-y+z)
=(x-y+z)(x+y+z-x+y+z)
=2(x+y+z)(y+z)
=(x+y+z)(x-y+z)-(x-y-z)(x-y+z)
=(x-y+z)(x+y+z-x+y+z)
=2(x+y+z)(y+z)
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