Đáp án:
Giải thích các bước giải:
Ta có : {(x+√x2+2011)(x−√x2+2011)(y+√y2+2011)=2011(x−√x2+2011)(x+√x2+2011)(y+√y2+2011)(y−√y2−2011)=2011((y−√y2−2011)){(x+x2+2011)(x−x2+2011)(y+y2+2011)=2011(x−x2+2011)(x+x2+2011)(y+y2+2011)(y−y2−2011)=2011((y−y2−2011))
<=>{−2011(y+√y2+2011)=2011(x−√x2+2011)−2011(x+√x2+2011)=2011((y−√y2−2011))<=>{−2011(y+y2+2011)=2011(x−x2+2011)−2011(x+x2+2011)=2011((y−y2−2011))
<=>{−(y+√y2+2011)=(x−√x2+2011)−(x+√x2+2011)=(y−√y2−2011)<=>{−(y+y2+2011)=(x−x2+2011)−(x+x2+2011)=(y−y2−2011)
Cộng 2 phương trình trên ta có 2(x+y)=0 => x=-y => x2011+y2011=x−y
