Question

已知(x^2+1)(y^2+1)=4xy,求代数式x^2-3y+1的值

x^2是x的平方

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

(X^2+1)(Y^2+1)=4XY
x^2y^2+(x^2+y^2)+1-4xy=0
(x^2y^2-2xy+1)+(x^2-2xy+y^2)=0
(xy-1)^2+(x-y)^2=0
平方数非负
要使等式成立
只能
xy-1=0
x-y=0
x=1,y=1
或者x=-1,y=-1
X^2-3Y+1
=1-3+1
=-1
或者
X^2-3Y+1
=1+3+1
=5