Решите систему уравнений

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

$\left \{ {{0.5x^2+xy+0.5y^2-x-y-4=0} \atop {0.5x^2+xy+0.25y^2+2=0}} \right.$
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: $(4+2 \sqrt{10};-2 \sqrt{10}),\,\,\,(4-2 \sqrt{10};2 \sqrt{10}),\,\,\,(2;-4),\,\,\,(-6;4)$