Question

解方程(x2+1)*(y2+1)=10、(x+y)*(xy-1)=3

Answer 1

(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