Question

Нужна помощь с 13 срочно

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Answer 1

Ответ:

$\left\{\begin{array}{l}x(1+y)+4=-4y\\x^2+y^2=5\end{array}\right\ \ \left\{\begin{array}{l}x(1+y)+4(1+y)=0\\x^2+y^2=5\end{array}\right\ \ \left\{\begin{array}{l}(1+y)(x+4)=0\\x^2+y^2=5\end{array}\right$

$\left\{\begin{array}{l}1+y=0\ ,\ x+4=0\\x^2+y^2=5\end{array}\right\ \ \left\{\begin{array}{l}y_1=-1\ ,\ x_2=-4\\x^2+y^2=5\end{array}\right\\\\\\a)\ \ \left\{\begin{array}{l}y=-1\\x^2+1=5\end{array}\right\ \ \left\{\begin{array}{l}y=-1\\x^2=4\end{array}\right\ \ \left\{\begin{array}{l}y=-1\\x_1=-2\ ,\ x_2=2\end{array}\right\ \ \ \ (-2;-1)\ ,\ (2;-1)$

$b)\ \ \left\{\begin{array}{l}x=-4\\16+y^2=5\end{array}\right\ \ \left\{\begin{array}{l}x=-4\\y^2=-11\end{array}\right\ \ \left\{\begin{array}{l}x=-4\\y\in \varnothing \end{array}\right\\\\\\Otvet:\ \ (-2\, ;-1)\ ,\ (2\, ;-1)\ .$

Answer 2

$\displaystyle\bf\\\left \{ {{x(1+y)+4=-4y} \atop {x^{2} +y^{2}=5 }} \right. \\\\\\\left \{ {{x(1+y)+(4+4y)=0} \atop {x^{2} +y^{2}=5 }} \right. \\\\\\\left \{ {{x(1+y)+4(1+y)=0} \atop {x^{2} +y^{2}=5 }} \right. \\\\\\\left \{ {{(1+y)\cdot (x+4)=0} \atop {x^{2} +y^{2}=5 }} \right. \\\\\\(1+y)\cdot (x+4)=0\\\\\\\left \{ {{1+y=0} \atop {x+4=0}} \right. \\\\\\\left \{ {{y=-1} \atop {x=-4}} \right. \\\\\\x^{2} +y^{2} =5\\\\y=-1 \ \ \Rightarrow x^{2} =4 \ \ ; \ x=\pm 2$

$\displaystyle\bf\\x=-4 \ \ \Rightarrow \ \ y^{2} =-11 \ \ ; \ y\in \oslash\\\\\\Otvet:(-2 \ ; \ -1) \ , \ (2 \ ; - 1)$