Question

Решите под Б пожалуйста

Answer 1

$left { {{x^2y^2-xy=12} atop {x+y=2}} right. \ left { {{x^2y^2-xy-12=0} atop {x+y=2}} right. \ left { {{(xy-4)(xy+3)=0} atop {x+y=2}} right.$

$left { {{xy-4=0} atop {x+y=2}} right. \ x=2-y \ (2-y)y-4=0 \ y^2-2y+4=0 \ D_1=1-4<0$

$left { {{xy+3=0} atop {x+y=2}} right. \ x=2-y \ (2-y)y+3=0 \ y^2-2y-3=0 \ D_1=1+3=4 \ y_1=1+2=3 \ y_2=1-2=-1 \ x_1=2-3=-1 \ x_2=2+1=3$

Ответ: (-1; 3); (3; -1)