若实数x、y满足x^2+y^2+2xy+x-y=0,则x(max)=____,y(min)=____.

问题描述:

若实数x、y满足x^2+y^2+2xy+x-y=0,则x(max)=____,y(min)=____.

?题对?

x^2+(2y+1)x+y^2-y=0,
△=(2y+1)^2-4(y^2-y)>=0
8y>=-1
y>=-1/8
y^2+(2x-1)y+x^2+x=0,
△=(2x-1)^2-4(x^2+x)>=0
x<=1/8
所以,x(max)=_1/8,y(min)=_-1/8