解方程(x2+1)*(y2+1)=10、(x+y)*(xy-1)=3

1个回答

  • (x²+1)(y²+1)=10

    x²y²+x²+y²+1=10

    x²y²-2xy+1+x²+y²+2xy=10

    (xy-1)²+(x+y)²=10

    [(xy-1)+(x+y)]²-2(x+y)(xy-1)=10

    (x+y)(xy-1)=3代入

    [(xy-1)+(x+y)]²=16

    (xy-1)+(x+y)=4或(xy-1)+(x+y)=-4

    x+y和xy-1是方程t²-4t+3=0或t²+4t+3=0的根.

    t²-4t+3=0 (t-1)(t-3)=0 t=1或t=3

    x+y=1 xy-1=3 xy=4 x,y是方程m²-m+4=0的根,方程无实根.

    x+y=3 xy-1=1 xy=2 x,y是方程m²-3m+2=0的根.

    (m-1)(m-2)=0 m=1或m=2

    x=1 y=2或x=2 y=1

    t²+4t+3=0 (t+1)(t+3)=0 t=-1或t=-3

    x+y=-1 xy-1=-3 xy=-2 x,y是方程m²+m-2=0的根,

    (m-2)(m+1)=0 m=2或m=-1

    x=2 y=-1或x=-1 y=2

    x+y=-3 xy-1=-1 xy=0 x,y是方程m²+3m=0的根.

    m(m+3)=0 m=0或m=-3

    x=0 y=-3或x=-3 y=0

    综上,得:

    x=1 y=2

    x=2 y=1

    x=2 y=-1

    x=-1 y=2

    x=0 y=-3

    x=-3 y=0