Question

**РЕШИТЕ СИСТЕМУ УРАВНЕНИЙ ****((ЖЕЛАТЕЛЬНО ПОДРОБНО ДАЮ 100 балов
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Answer 1

$\left \{ {{\frac{x^{2} }{y}+\frac{y^{2} }{x}=12} \atop {\frac{1}{x}+\frac{1}{y}=\frac{1}{3}}} \right.\\\\\left \{ {{\frac{x^{3}+y^{3}}{xy}=12 } \atop {\frac{x+y}{xy}=\frac{1}{3}}} \right.\\\\\left \{ {{\frac{(x+y)(x^{2}-xy+y^{2})}{xy}=12 } \atop {xy=3(x+y)}} \right.\\\\\left \{ {{\frac{(x+y)(x^{2}-xy+y^{2})}{3(x+y)}=12 } \atop {xy=3(x+y)}} \right.\\\\\left \{ {{x^{2}-xy+y^{2}=36} \atop {xy=3(x+y)}} \right.\\\\\left \{ {{(x+y)^{2}-3xy=36 } \atop {xy=3(x+y)}} \right.$

$\left \{ {{(x+y)^{2}-3*3(x+y)=36 } \atop {xy=3(x+y)}} \right. \\\\\left \{ {{(x+y)^{2}-9(x+y)-36=0 } \atop {xy=3(x+y)}} \right. \\\\x+y=m\\\\m^{2}-9m-36=0\\\\m_{1}=12\\\\m_{2}=-3\\\\1)\left \{ {{x+y=12} \atop {xy=3*12}} \right.\\\\\left \{ {{x+y=12} \atop {xy=36}} \right.\\\\\left \{ {{x=6} \atop {y=6}} \right.\\\\2)\left \{ {{x+y=-3} \atop {xy=3*(-3)}} \right. \\\\\left \{ {{x+y=-3} \atop {xy=-9}} \right. \\\\\left \{ {{x=-y-3} \atop {(-y-3)*y+9=0}} \right.$

$\left \{ {{x=-y-3} \atop {y^{2}+3y-9=0 }} \right. \\\\y^{2}+3y-9=0\\\\D=3^{2}-4*(-9)=9+36=45=(3\sqrt{5})^{2}\\\\y_{1} =\frac{-3-3\sqrt{5} }{2} \\\\y_{2}=\frac{-3+3\sqrt{5} }{2}\\\\x_{1}= -\frac{-3-3\sqrt{5} }{2} -3=\frac{3+3\sqrt{5}-6 }{2}=\frac{-3+3\sqrt{5}}{2}\\\\x_{2}=-\frac{-3+3\sqrt{5} }{2}-3=\frac{3-3\sqrt{5}-6 }{2}=\frac{-3-3\sqrt{5}}{2}$

$Otvet:\boxed{(6 \ ; \ 6) \ , \ (\frac{-3+3\sqrt{5} }{2} \ ; \ \frac{-3-3\sqrt{5}}{2}) \ , \ (\frac{-3-3\sqrt{5} }{2} \ ; \ \frac{-3+3\sqrt{5} }{2} )}$