Question

решите срочно!!!!!!!!!!!

Answer 1

Ответ:

решение смотри на фотографии

Answer 2

Решение.

Выразим  ху  из 2 уравнения и подставим в 1 уравнение .

$\left\{\begin{array}{l}\bf (x+2)(y+2)\, xy=9\\\bf x-xy+y=1\end{array}\right\ \ \left\{\begin{array}{l}\bf (xy+2x+2y+4)\, xy=9\\\bf xy=x+y-1\end{array}\right\ \ \\\\\\\left\{\begin{array}{l}\bf ((x+y-1)+2x+2y+4)\, xy=9\\\bf xy=x+y-1\end{array}\right\ \ \left\{\begin{array}{l}\bf (3x+3y+3)(x+y-1)=9\\\bf xy=x+y-1\end{array}\right$

Сначала вынесем множитель 3 за скобки, а потом воспользуемся формулой разности квадратов .

$\left\{\begin{array}{l}\bf 3\, ((x+y)+1)((x+y)-1)=9\\\bf xy=x+y-1\end{array}\right\ \ \left\{\begin{array}{l}\bf (x+y)^2-1^2=3\\\bf xy=x+y-1\end{array}\right\ \ \\\\\\\left\{\begin{array}{l}\bf x^2+2xy+y^2-1=3\\\bf x+y=xy+1\end{array}\right\ \ \left\{\begin{array}{l}\bf x^2+2xy+y^2=4\\\bf x+y=xy+1\end{array}\right\ \ \left\{\begin{array}{l}\bf (x+y)^2=4\\\bf x+y=xy+1\end{array}\right$

$\left\{\begin{array}{l}\bf (x+y=\pm 2\\\bf x+y=xy+1\end{array}\right\ \ \ \Rightarrow$  

Теперь решаем две системы уравнений .

$a)\ \left\{\begin{array}{l}\bf x+y=-2\\\bf x+y=xy+1\end{array}\right\ \ \ \ \ ili\ \ \ \ \ \ b)\ \left\{\begin{array}{l}\bf x+y=2\\\bf x+y=xy+1\end{array}\right$

$a)\ \left\{\begin{array}{l}\bf x+y=-2\\\bf -2=xy+1\end{array}\right\ \ \left\{\begin{array}{l}\bf x+y=-2\\\bf xy=-3\end{array}\right\ \ \left\{\begin{array}{l}\bf y=-x-2\\\bf x(-x-2)=-3\end{array}\right\ \ \left\{\begin{array}{l}\bf y=-x-2\\\bf x^2+2x-3=0\end{array}\right\\\\\\\left\{\begin{array}{l}\bf y=-x-2\\\bf x_1=-3\ ,\ x_2=1\ (Viet)\end{array}\right\ \ \left\{\begin{array}{l}\bf y_1=1\ ,\ y_2=-3\\\bf x_1=-3\ ,\ x_2=1\end{array}\right\ \ \ \boldsymbol{(-3;1)\ ,\ \bf (1;-3)}$  

$b)\ \left\{\begin{array}{l}\bf x+y=2\\\bf 2=xy+1\end{array}\right\ \ \left\{\begin{array}{l}\bf x+y=2\\\bf xy=1\end{array}\right\ \ \left\{\begin{array}{l}\bf y=-x+2\\\bf x(-x+2)=1\end{array}\right\ \ \left\{\begin{array}{l}\bf y=-x+2\\\bf x^2-2x+1=0\end{array}\right\\\\\\\left\{\begin{array}{l}\bf y=-x+2\\\bf (x-1)^2=0\end{array}\right\ \ \left\{\begin{array}{l}\bf y=1\\\bf x=1\end{array}\right\ \ \ \ \ \ \ \boldsymbol{(\, 1\, ;\, 1\, )}$  

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