0 (2 ,0 ĐIỂ M) + = + = + =3 3 3 ⇔ ⇔1 2 1 2 1 2X Y X Y X Y TA CÓ
1,0
(2 ,0 điể m)
+ = + = + =3
3
3
⇔ ⇔1 2 1 2 1 2x y x y x y Ta có:
+ = − = − − − + + = 3
3
3
2
2
y x x y y x x y x xy y1 2 2( ) ( )( 2) 0
0,25
2
2
y y2
2
3Do
2 2 0 ,x −xy+y + =x− + + > ∀x y2 4 + =
3
x y1 20,5
⇒ + = ⇔ − + − = =3
2