Question

система{x+y+xy=0{x^3+y^3+x^3y^3=12

Answer 1

$\left \{ {{x+y+xy=0} \atop {x^3+y^3+x^3y^3=12}} \right. \; \; \left \{ {{xy=-x-y} \atop {x^3+y^3+(-x-y)^3=12}} \right. \; ,\; \; xy=-(x+y)\; ,\; x+y=-xy\\\\\\x^3+y^3+(-x-y)^3=x^3+y^3-(x+y)^3=\\\\=x^3+y^3-(x^3+y^3+3x^2y+3xy^2)=-3x^2y-3xy^2=\\\\=-3xy(x+y)=3\cdot (-xy)(x+y)=3(x+y)\cdot (x+y)=3(x+y)^2\\\\3(x+y)^2=12\\\\(x+y)^2=4\quad \Rightarrow \; \; \; x+y=\pm 2\\\\1)\; \; x+y=-2\; \; \Rightarrow \; \; \; y=-2-x\\\\x+y+xy=x+(-2-x)+x(-2-x)=-2-2x-x^2=0\\\\x^2+2x+2=0\; ,\; \; D=4-8=-4\ \textless \ 0\; \; net\; kornej$

$2)\; \; x+y=2\; \; \Rightarrow \; \; \; y=2-x\\\\x+y+xy=x+(2-x)+x(2-x)=2+2x-x^2=0\\\\x^2-2x-2=0\; ,\; \; D=4+8=12\; ,\\\\x_1= \frac{2-2\sqrt{3}}{2} =1-\sqrt3\; ,\; \; x_2=1+\sqrt3\\\\y_1=2-(1-\sqrt3)=1+\sqrt3\; ,\; \; y_2=2-(1+\sqrt3)=1-\sqrt3\\\\Otvet:\; \; (1-\sqrt3\; ;\; 1+\sqrt3)\; ,\; \; (1+\sqrt3\; ;\; 1-\sqrt3)\; .$