Question

помагите пожалуйста ​ ​

Answer 1

Ответ:

$\left\{\begin{array}{l}y^2-xy=12\\x^2-xy=-3\end{array}\right\ \ \left\{\begin{array}{l}y(y-x)=12\\x(x-y)=-3\end{array}\right\ \ \left\{\begin{array}{l}x-y=-\dfrac{12}{y}\\x-y=-\dfrac{3}{x}\end{array}\right\\\\\\-\dfrac{12}{y}=-\dfrac{3}{x}\ \ ,\ \ \ \dfrac{y}{x}=4\ \ ,\ \ y=4x$

$\left\{\begin{array}{l}x-y=-\dfrac{12}{y}\\x-4x=-\dfrac{3}{x}\end{array}\right\ \ \left\{\begin{array}{l}x-y=-\dfrac{12}{y}\\-3x^2=-3\end{array}\right\ \ \left\{\begin{array}{l}x-y=-\dfrac{12}{y}\\x^2=1\end{array}\right\ \ \left\{\begin{array}{l}x-y=-\dfrac{12}{y}\\x=\pm 1\end{array}\right$

$a)\ \ \left\{\begin{array}{l}x-y=-\dfrac{12}{y}\\x=-1\end{array}\right\ \ \left\{\begin{array}{l}-1-y=-\dfrac{12}{y}\\x=-1\end{array}\right\ \ \left\{\begin{array}{l}y^2+y-12=0\\x=-1\end{array}\right\\\\\\\left\{\begin{array}{l}y_1=-4\ ,\ y_2=3\\x=-1\end{array}\right\ \ \ \Rightarrow \ \ \ (-1;-4)\ ,\ (-1;\ 3\ )\ \ -\ otvet$

$b)\ \ \left\{\begin{array}{l}x-y=-\dfrac{12}{y}\\x=1\end{array}\right\ \ \left\{\begin{array}{l}1-y=-\dfrac{12}{y}\\x=1\end{array}\right\ \ \left\{\begin{array}{l}y^2-y-12=0\\x=1\end{array}\right\\\\\\\left\{\begin{array}{l}y_1=-3\ ,\ y_2=4\\x=1\end{array}\right\ \ \ \Rightarrow \ \ \ (1;-3)\ ,\ (1;\, 4\, )\ \ -\ otvet$

$Otvet:\ \ (-1;-4\, )\ ,\ (-1\ ;\ 3\ )\ ,\ (\ 1\ ;-3\ )\ ,\ (\ 1\ ;\ 4\ )\ .$