Question

помогите пожалуйста номер 14.5​

Answer 1

Ответ:

$1)\ \ \left\{\begin{array}{l}\sqrt{x}+\sqrt{y} =3\\\sqrt{x}-\sqrt{y}=1\end{array}\right\ \oplus\ \ominus \ \left\{\begin{array}{l}2\sqrt{x}=4\\2\sqrt{y}=2\end{array}\right\ \ \left\{\begin{array}{l}\sqrt{x}=2\\\sqrt{y}=1\end{array}\right\ \ \left\{\begin{array}{l}x=4\\y=1\end{array}\right\\\\\\ODZ:\ x>0\ ,\ y>0\ .\ \ \ Otvet:\ \ (\, 4\ ;\ 1\ )\ .$

$2)\ \ \left\{\begin{array}{l}\sqrt{x}+\sqrt{y}=5\\x+y=13\end{array}\right\ \ \left\{\begin{array}{l}\sqrt{x}+\sqrt{y}=5\\(\sqrt{x}+\sqrt{y})^2-2\sqrt{xy}=13\end{array}\right\ \ \left\{\begin{array}{l}\sqrt{x}+\sqrt{y}=5\\25-2\sqrt{xy}=13\end{array}\right$

$\left\{\begin{array}{l}\sqrt{x}+\sqrt{y}=5\\2\sqrt{xy}=12\end{array}\right\ \ \left\{\begin{array}{l}\sqrt{y}=5-\sqrt{x}\\\sqrt{x}\cdot (5-\sqrt{x})=6\end{array}\right\ \ \left\{\begin{array}{l}\sqrt{y}=5-\sqrt{x}\\x-5\sqrt{x}+6=0\end{array}\right\\\\\\x-5\sqrt{x}+8=0\ \ ,\ \ t=\sqrt{x} \geq 0\ \ ,\ \ t^2-5t+6=0\ \ ,\ \ t_1=2\ ,\ t_2=3$

$\left\{\begin{array}{l}\sqrt{y}=5-\sqrt{x}\\\sqrt{x}=2\ ,\ \sqrt{x}=3\end{array}\right\ \ \left\{\begin{array}{l}\sqrt{y}=3\ ,\ \sqrt{y}=2\\\sqrt{x}=2\ ,\ \sqrt{x}=3\end{array}\right\ \ \left\{\begin{array}{l}y_1=9\ ,\ y_2=4\\x_1=4\ ,\ x_2=9\end{array}\right\\\\\\ODZ:\ x>0\ ,\ y>0\ .\ \ \ \ \ Otvet:\ (\, 4\, ;\, 9\, )\ ,\ (\, 9\, ;\, 4\, )\ .$