Question

помогите всё на фото ​

Answer 1

Ответ:

Метод сложения решения систем уравнений .

Вычтем из 1-го уравнения 2-ое .

$\left\{\begin{array}{l}x^2-y^2=8\\x^2+2y^2=32\end{array}\right\ \ominus \ \left\{\begin{array}{l}x^2-y^2=8\\-3y^2=-24\end{array}\right\ \ \left\{\begin{array}{l}x^2=y^2+8\\y^2=8\end{array}\right\ \ \left\{\begin{array}{l}x^2=y^2+8\\y_{1,2}=\pm \sqrt8\end{array}\right$

$a)\ \ \left\{\begin{array}{l}x^2=8+8\\y_{1}=-2\sqrt2\end{array}\right\ \ \left\{\begin{array}{l}x^2=16\\y_{1}=-2\sqrt2\end{array}\right\ \ \left\{\begin{array}{l}x_{1,2}=\pm 4\\y_{1}=-2\sqrt2\end{array}\right\\\\\\(-4;-2\sqrt2)\ ,\ (4;-2\sqrt2)$

$a)\ \ \left\{\begin{array}{l}x^2=8+8\\y_{1}=2\sqrt2\end{array}\right\ \ \left\{\begin{array}{l}x^2=16\\y_{1}=2\sqrt2\end{array}\right\ \ \left\{\begin{array}{l}x_{1,2}=\pm 4\\y_{1}=2\sqrt2\end{array}\right\\\\\\(-4;2\sqrt2)\ ,\ (4;2\sqrt2)\\\\\\Otvet:\ (-4;-2\sqrt2)\ ,\ (4;-2\sqrt2)\ ,\ (-4;2\sqrt2)\ ,\ (4;2\sqrt2)\\\\{}\ \ \ \ \ ili\ \ \ (-4;-\sqrt8)\ ,\ (4;-\sqrt8)\ ,\ (-4;\sqrt8)\ ,\ (4;\sqrt8)\ .$