Question

已知（x2+1）(y2+1)=4xy,求代数式x2--3y+1

Answer 1

(x^2+1)(y^2+1)=4xy
x^2y^2+x^2+y^2+1-4xy=0
(x^2y^2-2xy+1)+(x^2+y^2-2xy)=0
(xy-1)^2+(x-y)^2=0
(xy-1)^2=0,(x-y)^2=0
xy-1=0
x-y=0
解得x=y=1或x=y=-1
当x=y=1时,x^2-3y+1=1-3+1=-1
当x=y=-1时,x^2-3y+1=1+3+1=5
所以x^2-3y+1=-1或5

Answer 2

x²y²+x²+y²+1=2xy+2xy
(x²y²-2xy+1)+(x²-2xy+y²)=0
(xy-1)²+(x-y)²=0
所以xy-1=0,x-y=0
所以x=y=1或x=y=-1

所以x²-3y+1=1-3+1=-1
或x²-3y+1=1+3+1=5

Answer 3

x^2y^2+x^2+y^2+1-4xy=0
(xy-1)^2+(x-y)^2=0
∴xy=1,x=y.
∴x=y=1.
∴原式=1-3+1=-1