设方程组x³-xyz=-5,y³-xyz=2,z³-xyz=21的正实数解为(x,y,z)则x+y+z=
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设方程组x³-xyz=-5,y³-xyz=2,z³-xyz=21的正实数解为(x,y,z)则x+y+z=
设方程组x³-xyz=-5,y³-xyz=2,z³-xyz=21的正实数解为(x,y,z)则x+y+z=
设方程组x³-xyz=-5,y³-xyz=2,z³-xyz=21的正实数解为(x,y,z)则x+y+z=
我先把题目改的简单点,在正整数范围内解方程组
x³-xyz=-5,y³-xyz=2
两式相减,得到(y-x)(x^2+xy+y^2)=7
所以得到y-x=1或者y-x=7(舍去)
所以y-x=1,则x^2+xy+y^2=7
结合这个两个得到y=2,x=1
带入x³-xyz=-5中得到z=3
所以x+y+z=6
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