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已知y-2x=0,求代数式(x²-y²)(x²-xy+y²)/(x²+xy+y²)(x²-y²)的值

分类: 作业答案 • 2021-12-20 11:58:56
问题描述：

已知y-2x=0,求代数式(x²-y²)(x²-xy+y²)/(x²+xy+y²)(x²-y²)的值

答

y-2x=0
∴y=2x


(x²-y²)(x²-xy+y²)/(x²+xy+y²)(x²-y²)
=(x²-xy+y²)/(x²+xy+y²)
=(x²-2x²+4x²)/(x²+2x²+4x²)
=3x²/7x²
=3/7

答

即y=2x
所以原式=(x²-xy+y²)/(x²+xy+y²)
=(x²-2x²+4x²)/(x²+2x²+4x²)
=3x²/7x²
=3/7

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