已知y-2x=0,求代数式(x²-y²)(x²-xy+y²)/(x²+xy+y²)(x²-y²)的值
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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
即y=2x
所以原式=(x²-xy+y²)/(x²+xy+y²)
=(x²-2x²+4x²)/(x²+2x²+4x²)
=3x²/7x²
=3/7
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