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已知x²-xy-y²=0,且x≠0,y≠0.求代数式(x²-2xy-5y²)/(x²+2xy+5y²)的值
人气:467 ℃ 时间:2019-10-19 21:52:43
解答
(x+y)(x²-xy-y²)=0, => x^3-y^3=0, x=y
(x²-2xy-5y²)/(x²+2xy+5y²)=-6/8=-3/4=-0.75
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