Question

Решить систему неравенств
(x-y)^2-x+y=0
x^2y^2-xy-2=0

Answer 1

4дробь √11= Иии √5Дробь √5-2

Answer 2

$left { {{(x-y)^2-x+y=0,} atop {x^2y^2-xy-2=0,}} right. left { {{(x-y)^2-(x-y)=0,} atop {(xy)^2-xy-2=0,}} right. \ x-y=a, xy=b; \ a^2-a=0, \ a(a-1)=0, \ a_1=0, \ a-1=0, a_2=1; \ b^2-b-2=0, \ b_1=-1, b_2=2; \ left { {{ left [ {{x-y=0,} atop {x-y=1;}} right. } atop { left [ {{xy=-1,} atop {xy=-2;}} right. }} right. \ 1) left { {{x=y} atop {xy=-1;}} right. left { {{x=y} atop {y^2=-1;}} right. \ y^2=-1<0, \ yinvarnothing;$
$2) left { {{x=y,} atop {xy=2;}} right. left { {{x=y,} atop {y^2=2;}} right. \ y^2=2, \ y_1=- sqrt{2} , y_2= sqrt{2} , \ x_1=- sqrt{2} , x_2= sqrt{2};\ 3) left { {{x-y=1,} atop {xy=-1;}} right. left { {{y=x-1,} atop {x(x-1)=-1;}} right. \ x^2-x+1=0, \ D=-3<0, \ xinvarnothing; \ 4) left { {{x-y=1,} atop {xy=2;}} right. left { {{y=x-1,} atop {x(x-1)=2;}} right. \ x^2-x-2=0, \ x_3=-1, x_4=2, \ y_3=-2, y_4=1.$