Question

Решите систему уравнений
$0.5(x+y)^2-x-y=4$
$0.5x^2+xy+0.25y^2=-2$

Answer 1

$left { {{0.5(x+y)^2-x-y=4} atop {0.5x^2+xy+0.25y^2=-2}} right.$

$left { {{0.5x^2+xy+0.5y^2-x-y-4=0} atop {0.5x^2+xy+0.25y^2+2=0}} right.$
Разложим на множители первое уравнение
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$0,5(x^2+2xy+y^2-2x-2y-8)=0 \ 0.5(x^2+xy-4x+xy+y^2-4y+2x-4y+2y-8)=0 \ 0.5(x+y-4)(x+y+2)=0$
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Имеем 2 системы
$left { {{x+y-4=0} atop {0.5x^2+xy+0.25y^2+2=0}} right. to left { {{x=4-y} atop {(0.5(4-y)^2+(4-y)y+0.25y^2+2=0}} right. \ 0.25y^2=10 \ y^2=40 \ y=pm2 sqrt{10} \ x=pm4+2 sqrt{10}$

и 

$left { {{x+y+2=0} atop {0.5x^2+xy+0.25y^2+2=0}} right. to left { {{x=-y-2} atop {0.5(-y-2)^2+(-y-2)y+0.25y^2+2=0}} right.$
$0.25y^2=4 \ y^2=16 \ y=pm4 \ x=_1=2,,,,,x_2=-6$

Ответ: $(4+2 sqrt{10};-2 sqrt{10}),,,,(4-2 sqrt{10};2 sqrt{10}),,,,(2;-4),,,,(-6;4)$