x+y=x^2-xy+y^2+1 ===> x^2 - (y+1)x + y^2 - y + 1 = 0 ,x有实数解,要求：(y+1)^2 - 4*(y^2 - y + 1) = -3y^2 + 6y - 3 >= 0 .(1) -3y^2 + 6y - 3 = -3 * (y-1)^2只有 y = 1 ,满足 (1) and (2).所以,y = 1 ,代入,...