已知x^3+xy^2+x+x^2y+y^3+y=0,求5x-6y/3x+4y的值.
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已知x^3+xy^2+x+x^2y+y^3+y=0,求5x-6y/3x+4y的值.
(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)=0
x²+y²+1≥1>0
即不会等于0
所以x+y=0
y=-x
原式=(5x+6x)/(3x-4x)
=11x/(-x)
=-11
(x+y)(x²-xy+y²)+xy(x+y)+(x+y)=0
(x+y)(x²-xy+y²+xy+1)=0
(x+y)(x²+y²+1)=0
x²+y²+1≥1>0
即不会等于0
所以x+y=0
y=-x
原式=(5x+6x)/(3x-4x)
=11x/(-x)
=-11
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