Question

х2+у2=4 у+х=0 срочна пжолуста ​

Answer 1

$\displaystyle\left \{ {{x^{2}+y^{2}=4 } \atop {y+x=0}} \right.\\\\\\\displaystyle\left \{ {{x^{2}+(-x)^{2}=4 } \atop {y=-x}} \right.\\\\\\\displaystyle\left \{ {{x^{2}+x^{2}=4 } \atop {y=-x}} \right.\\\\\\\displaystyle\left \{ {{2x^{2}=4 } \atop {y=-x}} \right.\\\\\\\displaystyle\left \{ {{x^{2}=2 } \atop {y=-x}} \right.\\\\\\\displaystyle\left \{ {{x_{1,2} =\pm \sqrt{2}} \atop {y=-x}} \right.$

$\left[\begin{array}{ccc}\left \{ {{x_{1} =-\sqrt{2} } \atop {y_{1}=\sqrt{2}}} \right. \\\\\left \{ {{x_{2}=\sqrt{2}} \atop {y_{2}=-\sqrt{2}}} \right. \end{array}\right\\\\\\Otvet:\boxed{(-\sqrt{2} \ ; \ \sqrt{2} ) \ ; \ (\sqrt{2} \ ; \ -\sqrt{2})}$