Question

已知x²+y²+4x-8y+20=0，则x为多少，y为多少；（x²+y²）²-y²=x²+6，则

Answer 1

(x+2)^2+(y-4)^2=0
x=-2,y=4
(x^2+y^2)^2=y^2+x^2+6=(x+y)^2
2xy=6,xy=3,x^2+y^2=x+y
(3/y)^2+y^2=3/y+y
3+y^4=3y+y^3
y^3(y-1)=3(y-1)
y=3^(1/3),x=3^(2/3)

Answer 2

x²+y²+4x-8y+20=0
则(x+2)^2+(y-4)^2=0
两个因子都是非负数。要使得等式成立
则
x+2=0
y-4=0
x=-2 y=4