Question

помогите пожалуйста, кто шарит

Answer 1

Ответ:

$(4; 2\sqrt{2} ),(4; -2\sqrt{2} ),(-4; 2\sqrt{2} ),(-4; -2\sqrt{2} )$

Пошаговое объяснение:

$\left \{\begin{array}{l} x^{2} - y^{2} = 8, \\ 2x^{2} +2y^{2} =48|:2 ; \end{array} \right.\Leftrightarrow \left \{\begin{array}{l} x^{2} - y^{2} = 8, \\ x^{2} +y^{2} =24 ; \end{array} \right.\Leftrightarrow\left \{\begin{array}{l} 2x^{2} = 32, \\ x^{2} +y^{2} =24 ; \end{array} \right.\Leftrightarrow$

$\Leftrightarrow\left \{\begin{array}{l} x^{2} = 32:2, \\ x^{2} +y^{2} =24 ; \end{array} \right.\Leftrightarrow\left \{\begin{array}{l} x^{2} = 16, \\ 16 +y^{2} =24 ; \end{array} \right.\Leftrightarrow\left \{\begin{array}{l} x^{2} = 16, \\ y^{2} =8 ; \end{array} \right.\Leftrightarrow$

$\Leftrightarrow\left \{\begin{array}{l} \left [\begin{array}{l} x = -4 \\ x = 4;\end{array} \right., \\ \left [\begin{array}{l}y =- \sqrt{8} , \\ y = \sqrt{8}; \end{array} \right. \end{array} \right.\Leftrightarrow \left \{\begin{array}{l} \left [\begin{array}{l} x = -4 \\ x = 4;\end{array} \right., \\ \left [\begin{array}{l}y =- 2\sqrt{2} , \\ y = 2\sqrt{2}; \end{array} \right. \end{array} \right.\Leftrightarrow$

$\left [\begin{array}{l} \left \{\begin{array}{l} x= 4, \\ y =2\sqrt{2}; \end{array} \right. \\ \left \{\begin{array}{l} x = 4, \\ y = -2\sqrt{2} ; \end{array} \right. \\ \left \{\begin{array}{l} x = -4, \\ y = -2\sqrt{2} ; \end{array} \right.\\ \left \{\begin{array}{l} x = -4, \\ y = 2\sqrt{2} . \end{array} \right.\end{array} \right.$

Тогда система имеет четыре решения

$1. \left \{\begin{array}{l} x = 4, \\ y = 2\sqrt{2} \end{array} \right.$

$2. \left \{\begin{array}{l} x = 4, \\ y = -2\sqrt{2} \end{array} \right.$

$3. \left \{\begin{array}{l} x = - 4, \\ y = 2\sqrt{2} \end{array} \right.$

$4. \left \{\begin{array}{l} x =- 4, \\ y = -2\sqrt{2} \end{array} \right.$