Question

Заранее
решить систему уравнений
с обьяснением.
думаю, ответ (4;0)​

Answer 1

$\left\{\begin{array}{ccc}\sqrt{x^2-y^2}+\sqrt{x-y}=6\\x^2-y^2-x+y=12\end{array}\right\ \ \left\{\begin{array}{l}\sqrt{(x-y)(x+y)}+\sqrt{x-y}=6\\(x-y)(x+y)-(x-y)=12\end{array}\right\\\\\\u=\sqrt{x-y}\ ,\ \ v=\sqrt{x+y}\ \ \ \left\{\begin{array}{l}uv+u=6\\u^2v^2-u^2=12\end{array}\right\ \ \left\{\begin{array}{l}uv=6-u\\(6-u)^2-u^2=12\end{array}\right$

$\left\{\begin{array}{l}uv=6-u\\36-12u+u^2-u^2=12\end{array}\right\ \ \left\{\begin{array}{l}v=\dfrac{6-u}{u}\\12u=24\end{array}\right\ \ \left\{\begin{array}{ccc}v=2\\u=2\end{array}\right\ \ \left\{\begin{array}{ccc}\sqrt{x+y}=2\\\sqrt{x-y}=2\end{array}\right$

$\left\{\begin{array}{ccc}x+y=4\\x-y=4\end{array}\right\ \oplus \ \left\{\begin{array}{l}y=4-x\\2x=8\end{array}\right\ \ \left\{\begin{array}{l}y=0\\x=4\end{array}\right$

$ODZ:\ \ \left\{\begin{array}{l}x^2-y^2\geq 0\\x-y\geq 0\end{array}\right\ \ \left\{\begin{array}{l}x^2\geq y^2\\x\geq y\end{array}\right\ \ \ \Rightarrow \ \ \ \left\{\begin{array}{l}4^2\geq 0^2\ verno\\4\geq 0\ verno\end{array}\right\\\\\\Otvet:\ \ (\, 4\ ;\ 0\ )\ .$