Question

已知x^2+4y^2-6x+4y+10=0，求xy的值

Answer 1

解：x^2-6x+4y^2+4y+10=0
(x^2-6x+9-9)+(4y^2+4y)+10=0
(x-3)^2-9+4(y^2+y)+10=0
(x-3)^2-9+4x[(y^2+y+1/4-1/4]+10=0
(x-3)^2-9+4x[(y+1/2)^2-1/4]+10=0
(x-3)^2-9+4(y+1/返如行2)^2-1+10=0
(x-3)^2+4（y+1/2）^2=0
(x-3)^2>=0,4(y+1/2)^2>=0
(x-3)^2+4(y+1/橡樱2)^2>=0+0=0
(x-3)^2+4(y+1/2)^2>=0
当取到等号时漏哗，（x-3）^2=0且4（y+1/2）^2=0
x=3,y=-1/2
xy=3x(-1/2)=-3/2
答：答案是-3/2.

Answer 2

即(x²孝亮-6x+9)+(4y²+4y+1)=0
(x-3)²+(2y+1)²=0
所巧睁宽以早枯x-3=0,2y+1=0
x=3,y=-1/2
所以xy=-3/2