Question

Решите систему
x^2 + y^2 - 16 = 2xy
x^2 + y^2 - 4 = -2xy

Answer 1

$left { {{x^2+y^2-16=2xy} atop {x^2+y^2-4=-2xy}} right. ; left { {{x^2-2xy+y^2=16} atop {x^2+2xy+y^2=4}} right. ; left { {{(x-y)^2=16} atop {(x+y)^2=4}} right. ; left { {{x-y=pm 4} atop {x+y=pm 2}} right. \\\ left { {{x-y=4} atop {x+y=2}} right. ; ili; left { {{x-y=-4} atop {x+y=2}} right. ; ili; left { {{x-y=4} atop {x+y=-2}} right. ; ili; left { {{x-y=-4} atop {x+y=-2}} right.$

 $1); left { {{x-y=4} atop {2x=6}} right.$.$left { {{y=-1} atop {x=3}} right.$ 

$2); ; left { {{x-y=-4} atop {2x=-2}} right. ; left { {{y=3} atop {x=-1}} right. \\3); ; left { {{x-y=4} atop {2x=2}} right. ; left { {{y=-3} atop {x=1}} right. \\4); ; left { {{x-y=-4} atop {2x=-6}} right. ; left { {{y=1} atop {x=-3}} right. \\Otvet:; (3,-1); ,; (-1,3); ,; (1,-3); ,; (-3,1); .$

Answer 2

Спасибо огромное!