Question

Даю 100 балов

Розв'яжіть системи рівнянь

Answer 1

Ответ:

Решить системы уравнений . Применяем замену переменной .

1)  

$\left\{\begin{array}{l}\bf \sqrt[4]{\bf x+y}+\sqrt[4]{\bf x-y}=6\\\bf \sqrt{x+y}-\sqrt{x-y}=12\end{array}\right\ \ \ Zamena:\ u=\sqrt[4]{\bf x+y}\ \ ,\ \ v=\sqrt[4]{\bf x-y}\\\\\\\left\{\begin{array}{l}\bf v+u=6\\\bf u^2-v^2=12\end{array}\right\ \ \left\{\begin{array}{l}\bf v=6-u\\\bf u^2-(6-u)^2=12\end{array}\right\ \ \left\{\begin{array}{l}\bf v=6-u\\\bf u^2-(36-12u+u^2)=12\end{array}\right$  

$\left\{\begin{array}{l}\bf v=6-u\\\bf 12u-36=12\end{array}\right\ \ \left\{\begin{array}{l}\bf v=6-u\\\bf u-3=1\end{array}\right\ \ \left\{\begin{array}{l}\bf v=6-u\\\bf u=4\end{array}\right\ \ \left\{\begin{array}{l}\bf v=2\\\bf u=4\end{array}\right$          

Обратная замена.

$\left\{\begin{array}{l}\bf \sqrt[4]{\bf x+y}=4\\\bf \sqrt[4]{\bf x-y}=2\end{array}\right\ \ \left\{\begin{array}{l}\bf x+y=4^4\\\bf x-y=2^4\end{array}\right\ \ \left\{\begin{array}{l}\bf x+y=256\\\bf x-y=16\end{array}\right$  

Сложим уравнения системы и вычтем из 1уравнения второе .

$\left\{\begin{array}{l}\bf 2x=272\\\bf 2y=240\end{array}\right\ \ \left\{\begin{array}{l}\bf x=136\\\bf y=120\end{array}\right\\\\\\\bf Otvet:\ (\ 136\ ;\ 120\ )$                

2)

$\left\{\begin{array}{l}\bf \sqrt[3]{\bf x-2y}+\sqrt[3]{\bf x+y}=1\\\bf \sqrt[3]{\bf x-2y}\cdot \sqrt[3]{\bf x+y}=-2\end{array}\right\ \ Zamena:\ \bf u=\sqrt[3]{\bf x-2y}\ ,\ \ v=\sqrt[3]{\bf x+y}$        

$\left\{\begin{array}{l}\bf u+v=1\\\bf uv=-2\end{array}\right\ \ \left\{\begin{array}{l}\bf u=1-v\\\bf (1-v)\cdot v=-2\end{array}\right\ \ \left\{\begin{array}{l}\bf u=1-v\\\bf v-v^2=-2\end{array}\right\ \ \left\{\begin{array}{l}\bf u=1-v\\\bf v^2-v-2=0\end{array}\right\\\\\\\left\{\begin{array}{l}\bf u=1-v\\\bf v_1=-1\ ,\ v_2=2\end{array}\right\ \ \left\{\begin{array}{l}\bf u_1=2\ ,\ u_2=-1\\\bf v_1=-1\ ,\ v_2=2\end{array}\right$      

Обратная замена .

$\bf a)\ \ \left\{\begin{array}{l}\bf \sqrt[3]{\bf x-2y}=2\\\bf \sqrt[3]{\bf x+y}=-1\end{array}\right\ \ \left\{\begin{array}{l}\bf x-2y=2^3\\\bf x+y=(-1)^3\end{array}\right\ \ \left\{\begin{array}{l}\bf x-2y=8\\\bf x+y=-1\end{array}\right$                

Вычтем из второго уравнения первое .      

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

$\bf b)\ \ \left\{\begin{array}{l}\bf \sqrt[3]{\bf x-2y}=-1\\\bf \sqrt[3]{\bf x+y}=2\end{array}\right\ \ \left\{\begin{array}{l}\bf x-2y=(-1)^3\\\bf x+y=2^3\end{array}\right\ \ \left\{\begin{array}{l}\bf x-2y=-1\\\bf x+y=8\end{array}\right$    

Вычтем из второго уравнения первое .  

$\left\{\begin{array}{l}\bf 3y=9\\\bf x+y=8\end{array}\right\ \ \left\{\begin{array}{l}\bf y=3\\\bf x+3=8\end{array}\right\ \ \left\{\begin{array}{l}\bf y=3\\\bf x=5\end{array}\right$  

Ответ:   $\bf (2;-3)\ ,\ (\ 5\ ;\ 3\ )$  .