Question

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

Answer 1

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