Question

решите систему уравнений

Answer 1

Ответ:

$(\sqrt{6\frac{1}{3}}; \sqrt{6\frac{1}{3}}); \quad (-\sqrt{6\frac{1}{3}}; -\sqrt{6\frac{1}{3}}); \quad (-1; -3); \quad (1; 3);$

Пошаговое объяснение:

ОДЗ:

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

Решение:

$\left \{ {{3x^{2}-4xy+y^{2}=0} \atop {x^{2}+2y^{2}=19}} \right. \Leftrightarrow \left \{ {{x^{2}+2x^{2}-2xy-2xy+y^{2}=0} \atop {x^{2}+2y^{2}=19}} \right. \Leftrightarrow \left \{ {{x^{2}-2xy+y^{2}+2x^{2}-2xy=0} \atop {x^{2}+2y^{2}=19}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{(x-y)^{2}+2x(x-y)=0} \atop {x^{2}+2y^{2}=19}} \right. \Leftrightarrow \left \{ {{(x-y)(x-y+2x)=0} \atop {x^{2}+2y^{2}=19}} \right. \Leftrightarrow \left \{ {{(x-y)(3x-y)=0} \atop {x^{2}+2y^{2}=19}} \right. \Leftrightarrow \left \{ {{x-y=0} \atop {x^{2}+2y^{2}=19}} \right. \vee$

$\vee \left \{ {{3x-y=0} \atop {x^{2}+2y^{2}=19}} \right. \Leftrightarrow \left \{ {{x=y} \atop {x^{2}+2y^{2}=19}} \right. \vee \left \{ {{y=3x} \atop {x^{2}+2y^{2}=19}} \right. \Leftrightarrow \left \{ {{x=y} \atop {y^{2}+2y^{2}=19}} \right. \vee \left \{ {{y=3x} \atop {x^{2}+2 \cdot (3x)^{2}=19}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{x=y} \atop {3y^{2}=19}} \right. \vee \left \{ {{y=3x} \atop {x^{2}+2 \cdot 9x^{2}=19}} \right. \Leftrightarrow \left \{ {{x=y} \atop {y^{2}=\frac{19}{3}}} \right. \vee \left \{ {{y=3x} \atop {x^{2}+18x^{2}=19}} \right. \Leftrightarrow \left \{ {{x=y} \atop {y= \pm \sqrt{6\frac{1}{3}}}} \right. \vee \left \{ {{y=3x} \atop {19x^{2}=19}} \right. \Leftrightarrow$

$\Leftrightarrow \left \{ {{x= \pm \sqrt{6\frac{1}{3}}} \atop {y= \pm \sqrt{6\frac{1}{3}}}} \right. \vee \left \{ {{y=3x} \atop {x^{2}=1}} \right. \Leftrightarrow \left \{ {{x= \pm \sqrt{6\frac{1}{3}}} \atop {y= \pm \sqrt{6\frac{1}{3}}}} \right. \vee \left \{ {{y=3x} \atop {x= \pm 1}} \right. \Leftrightarrow \left \{ {{x= \pm \sqrt{6\frac{1}{3}}} \atop {y= \pm \sqrt{6\frac{1}{3}}}} \right. \vee \left \{ {{y= \pm 3} \atop {x= \pm 1}} \right. ;$

$(\sqrt{6\frac{1}{3}}; \sqrt{6\frac{1}{3}}); \quad (-\sqrt{6\frac{1}{3}}; -\sqrt{6\frac{1}{3}}); \quad (-1; -3); \quad (1; 3);$