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已知x^3+xy^2+x+x^2y+y^3+y=0,求5x-6y/3x+4y的值.
人气:410 ℃ 时间:2019-10-05 12:29:38
解答
(x³+y³)+(x²y+xy²)+(x+y)=0(x+y)(x²-xy+y²)+xy(x+y)+(x+y)=0(x+y)(x²-xy+y²+xy+1)=0(x+y)(x²+y²+1)=0x²+y²+1≥1>0即不会等于0所以x+y=0y=-x原式=(5x+6...
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