Question

СРОООЧНО!!!ПОЖАЛУЙСТА

Answer 1

Ответ:

$(0;-2); (\sqrt{3};1)(-\sqrt{3};1)$

Объяснение:

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

Решим первое уравнение системы:

$x^{2} +(x^{2} -2)^{2} =4\\x^2+x^4-4x^2+4=4\\x^4+x^2-4x^2+4-4=0\\x^4-3x^2=0\\x^2(x^2-3)=0\\x^2=0 < = > x=0\\x^2-3=0 < = > x^2=3 < = > x_{1} =\sqrt{3} ; x_{2} =-\sqrt{3}$

$\left \{ {{x=0} \atop {y=0^2-2}} \right. \\\left \{ {{x=0} \atop {y=-2}} \right.$$\left \{ {{x=\sqrt{3} } \atop {y=\sqrt{3} ^2-2}} \right. \\\left \{ {{x=\sqrt{3} } \atop {y=3-2}} \right. \\\left \{ {{x=\sqrt{3} } \atop {y=1}} \right.$$\left \{ {{x=-\sqrt{3} } \atop {y=(-\sqrt{3} )^2-2}} \right. \\\left \{ {{x=-\sqrt{3} } \atop {y=3-2}} \right. \\\left \{ {{x=-\sqrt{3} } \atop {y=1}} \right.$

Ответ: $(0;-2); (\sqrt{3};1)(-\sqrt{3};1)$