Question

помогите решить систему уравнений пожалуйста​

Answer 1

Ответ:

$(2; \quad 1)$

Объяснение:

ОДЗ:

$x \neq 0, \quad y \neq 0, \quad x+2y \neq 0;$

Решение:

$4) \quad \left \{ {{x^{3}+8y^{3}=16} \atop {2xy(x+2y)=16}} \right. \Leftrightarrow \left \{ {{x^{3}+(2y)^{3}=16} \atop {2xy(x+2y)=16}} \right. \Leftrightarrow \left \{ {{(x+2y)(x^{2}-2xy+4y^{2})=16} \atop {2xy(x+2y)=16}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{(x+2y)(x^{2}-2xy+4y^{2})-2xy(x+2y)=16-16} \atop {2xy(x+2y)=16}} \right. \Leftrightarrow \left \{ {{(x+2y)(x^{2}-2xy+4y^{2}-2xy)=0} \atop {2xy(x+2y)=16}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{(x+2y)(x^{2}-4xy+4y^{2})=0} \atop {2xy(x+2y)=16}} \right. \Leftrightarrow \left \{ {{(x+2y)(x-2y)^{2}=0} \atop {2xy(x+2xy)=16}} \right. \Leftrightarrow \left \{ {{(x-2y)^{2}=0} \atop {2xy(x+2y)=16}} \right. \Leftrightarrow \left \{ {{x-2y=0} \atop {2xy(x+2y)=16}} \right. \Leftrightarrow$$\Leftrightarrow \left \{ {{x=2y} \atop {2xy(x+2y)=16}} \right. \Leftrightarrow \left \{ {{x=2y} \atop {2 \cdot 2y \cdot y \cdot (2y+2y)=16}} \right. \Leftrightarrow \left \{ {{x=2y} \atop {4y^{2} \cdot 4y=16}} \right. \Leftrightarrow \left \{ {{x=2y} \atop {16y^{3}=16}} \right. \Leftrightarrow\left \{ {{x=2y} \atop {y^{3}=1}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{x=2y} \atop {y=1}} \right. \Leftrightarrow \left \{ {{x=2} \atop {y=1}} \right. ;$

$(2; \quad 1);$