Từ giả thiết ta có
+ x2+y2+xy=(x+y)2−xy=1⇒(x+y)2=xy+1≥0⇒xy≥−1
+ x2+y2+xy=(x−y)2+3xy=1⇒(x−y)2=−3xy+1≥0⇒xy≤13
Vậy −1≤xy≤13 (1)
A=x2+y2−3xy=(x2+y2+xy)−4xy=1−4xy (2)
Từ (1); (2)⇒−13≤A≤5
+ maxA=5⇔xy=−1⇒x2+y2=2⇒(x; y)=(−1; 1); (1; −1)
+ minA=−13⇔xy=13⇒x2+y2=23⇒(x; y)=(−1√3; −1√3); (1√3; 1√3)