已知4x^2+y^2-4x+2y+2=0,则xy=?
问题描述:
已知4x^2+y^2-4x+2y+2=0,则xy=?
答
4x^2+y^2-4x+2y+2=0
4x^2-4x+1+y^2+2y+1=0
(2x-1)^2+(y+1)^2=0
(2x-1)^2=0,(y+1)^2=0
x=1/2
y=-1
xy=1/2*(-1)=-1/2
答
4x^2+y^2-4x+2y+2=0
(2x+1)^2+(y+1)^2=0
x=-1/2,y=-1
xy=1/2
答
4x²+y²-4x+2y+2
=(2x-1)²+(y+1)²
=0
2x-1=0 y+1=0
x=1/2 y=-1
xy=1/2*(-1)=-2分之1