Question

Помогите решить под 4)

Answer 1

$\left \{ {{x^{3}y^{3}+x^{2}y^{4}=12} \atop {x^{4}y^{2}+x^{3}y^{3} =24}} \right. \\\\:\left \{ {{x^{2}y^{3}(x+y)=12} \atop {x^{3}y^{2}(x+y)=24}} \right.\\--------\\\\\frac{y}{x}=\frac{1}{2}\\\\x=2y\\\\(2y)^{3}*y^{3}+(2y)^{2}*y^{4}=12\\\\8y^{3}*y^{3}+4y^{2}*y^{4}=12\\\\8y^{6}+4y^{6}=12\\\\12y^{6}=12\\\\y^{6} =1\\\\y_{1}=1\\\\y_{2}=-1\\\\x_{1}=2*1=2\\\\x_{2}=2*(-1)=-2$

Ответ : (2 ; 1) , ( - 2 ; - 1)