Question

{ (х-2)(у-2)ху=9
{ ху-х-у=3
нужно решить систему уравнений с полным решением и ответом​

Answer 1

Решение .

Решить систему уравнений .

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

Сделаем замену :  $\bf u=x+y\ ,\ v=xy$ . Получим систему

$\left\{\begin{array}{l}\bf (v-2u+4)\cdot v=9\\\bf v-u=3\end{array}\right\ \ \left\{\begin{array}{l}\bf v^2-2uv+4v=9\\\bf v-u=3\end{array}\right\ \ \left\{\begin{array}{l}\bf v^2-2(v-3)v+4v=9\\\bf u=v-3\end{array}\right\\\\\\\left\{\begin{array}{l}\bf v^2-2v^2+6v+4v=9\\\bf u=v-3\end{array}\right\ \ \left\{\begin{array}{l}\bf v^2-10v+9=0\\\bf u=v-3\end{array}\right\ \ \left\{\begin{array}{l}\bf v_1=1\ ,\ v_2=9\ (Viet)\\\bf u=v-3\end{array}\right$

$\left\{\begin{array}{l}\bf v_1=1\ ,\ v_2=9\\\bf u_1=-2\ ,\ u_2=6\end{array}\right$  

Перейдём к старым переменным .

$\bf a)\ \left\{\begin{array}{l}\bf v=1\\\bf u=-2\end{array}\right\ \ \left\{\begin{array}{l}\bf xy=1\\\bf x+y=-2\end{array}\right\ \ \left\{\begin{array}{l}\bf x(-x-2)=1\\\bf y=-x-2\end{array}\right\ \ \left\{\begin{array}{l}\bf -x^2-2x=1\\\bf y=-x-2\end{array}\right\\\\\\\left\{\begin{array}{l}\bf x^2+2x+1=0\\\bf y=-x-2\end{array}\right\ \ \left\{\begin{array}{l}\bf (x+1)^2=0\\\bf y=-x-2\end{array}\right\ \ \left\{\begin{array}{l}\bf x=-1\\\bf y=-1\end{array}\right\ \ \Rightarrow \ \ \ (-1\ ;-1\ )$  

$\bf b)\ \left\{\begin{array}{l}\bf v=9\\\bf u=6\end{array}\right\ \ \left\{\begin{array}{l}\bf xy=9\\\bf x+y=6\end{array}\right\ \ \left\{\begin{array}{l}\bf x(-x+6)=9\\\bf y=-x+6\end{array}\right\ \ \left\{\begin{array}{l}\bf -x^2+6x=9\\\bf y=-x-2\end{array}\right\\\\\\\left\{\begin{array}{l}\bf x^2-6x+9=0\\\bf y=-x+6\end{array}\right\ \ \left\{\begin{array}{l}\bf (x-3)^2=0\\\bf y=-x+6\end{array}\right\ \ \left\{\begin{array}{l}\bf x=3\\\bf y=3\end{array}\right\ \ \Rightarrow \ \ \ (\ 3\ ;\ 3\ )$  

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