Question

система рівнянь:
{xy/x+y + 4(x+y)/xy = 4 ,

{xy/x-y + x-y/xy = 2​

Answer 1

Ответ:

$\bigg (-4 \ ; \ 1\dfrac{1}{3} \bigg )$

Пошаговое объяснение:

$D(x \ , \ y): \ \begin{cases} x \pm y \neq 0 \\ x \neq 0 \\ y \neq 0 \end{cases} \Leftrightarrow \begin{equation} \begin{cases} x \neq \mp y \\ x \neq 0 \\ y \neq 0 \end{cases};$

$\displaystyle \left \{ {{\dfrac{xy}{x+y}+\dfrac{4(x+y)}{xy}=4} \atop {\dfrac{xy}{x-y}+\dfrac{x-y}{xy}=2}} \right. \ ;$

Введём замены:

$\dfrac{xy}{x+y}=t_{1} \Rightarrow \dfrac{4(x+y)}{xy}=\dfrac{4}{t_{1}} \ ;$

$\dfrac{xy}{x-y}=t_{2} \Rightarrow \dfrac{x-y}{xy}=\dfrac{1}{t_{2}} \ ;$

Перепишем уравнения с учётом замен:

$t_{1}+\dfrac{4}{t_{1}}=4 \Rightarrow \dfrac{t_{1}^{2}+4}{t_{1}}=4 \Rightarrow t_{1}^{2}+4=4t_{1} \Rightarrow t_{1}^{2}-4t_{1}+4=0 \Rightarrow (t_{1}-2)^{2}=0 \Rightarrow$

$\Rightarrow t_{1}-2=0 \Rightarrow t_{1}=2;$

$t_{2}+\dfrac{1}{t_{2}}=2 \Rightarrow \dfrac{t_{2}^{2}+1}{t_{2}}=2 \Rightarrow t_{2}^{2}+1=2t_{2} \Rightarrow t_{2}^{2}-2t_{2}+1=0 \Rightarrow (t_{2}-1)^{2}=0 \Rightarrow$

$\Rightarrow t_{2}-1=0 \Rightarrow t_{2}=1;$

Вернёмся к системе:

$\displaystyle \left \{ {{\dfrac{xy}{x+y}=2} \atop {\dfrac{xy}{x-y}=1}} \right. \Leftrightarrow \left \{ {{xy=2(x+y)} \atop {xy=x-y}} \right. \Leftrightarrow \left \{ {{x-y=2x+2y} \atop {xy=x-y}} \right. \Leftrightarrow \left \{ {{-x-3y=0} \atop {xy=x-y}} \right. \Leftrightarrow$

$\displaystyle \Leftrightarrow \left \{ {{x+3y=0} \atop {xy=x-y}} \right. \Leftrightarrow \left \{ {{x=-3y} \atop {-3y^{2}=-3y-y}} \right. \Leftrightarrow \left \{ {{x=-3y} \atop {3y^{2}-4y=0}} \right. \Leftrightarrow \left \{ {{x=-3y} \atop {y(3y-4)=0}} \right. \Leftrightarrow$

$\displaystyle \Leftrightarrow \left \{ {{x=-3y} \atop {y=0}} \right. \vee \left \{ {{x=-3y} \atop {3y-4=0}} \right. \Leftrightarrow \left \{ {{x=0} \atop {y=0}} \right. \vee \left \{ {{x=-3y} \atop {3y=4}} \right. \Leftrightarrow \left \{ {{x=0} \atop {y=0}} \right. \vee \left \{ {{x=-4} \atop {y=\dfrac{4}{3}}} \right. \Leftrightarrow$

$\displaystyle \Leftrightarrow \left \{ {{x=0} \atop {y=0}} \right. \vee \left \{ {{x=-4} \atop {y=1\dfrac{1}{3}}} \right. \ ;$

Первая пара корней не удовлетворяет ОДЗ.

$\bigg (-4 \ ; \ 1\dfrac{1}{3} \bigg ) \ ;$