Question

решите систему уравнений
$x - y = 11$
$\sqrt{x} + \sqrt{y} = 11$
​

Answer 1

Ответ:

$x = 36\\y = 25$

Пошаговое объяснение:

$\left \{ {{x-y = 11} \atop {\sqrt{x} + \sqrt{y} = 11}} \right. \Leftrightarrow \left \{ {{x = y + 11} \atop {\sqrt{x} + \sqrt{y} = 11}} \right.$

$\sqrt{y+11} + \sqrt{y} = 11$

$(\sqrt{y+11} + \sqrt{y})^2 = (11)^2, y > 0, y > -11$

$y+11 + 2\sqrt{y+11}\sqrt{y} + y = 121$

$2y + 2\sqrt{y+11}\sqrt{y} = 110$

$y + \sqrt{(y+11)*y} = 55\\y + \sqrt{y^2+11y} = 55\\55-y = -\sqrt{y^2+11y} \\(55-y)^2 = \sqrt{y^2+11y} \\55^2 - 110y + y^2 = y^2+11y\\55^2 = 121y\\\\y = \frac{55^2}{121}\\y = 25 > 0\\\\x = 25+11 = 36\\\\36 - 25 = 11\\6 + 5 = 11$

Answer 2

Ответ:

(36, 25)

Пошаговое объяснение:

$\left\{\begin{array}{c}x-y=11\\\sqrt{x}+\sqrt{y}=11\end{array}\right. => \left\{\begin{array}{c}(\sqrt{x}+\sqrt{y})(\sqrt{x}-\sqrt{y})=11\\\sqrt{x}+\sqrt{y}=11\end{array}\right. => \left\{\begin{array}{c}\sqrt{x}-\sqrt{y}=1\\\sqrt{x}+\sqrt{y}=11\end{array}\right.=>\\ \left\{\begin{array}{c}2\sqrt{x}=12\\2\sqrt{y}=10\end{array}\right.=>\left\{\begin{array}{c}\sqrt{x}=6\\\sqrt{y}=5\end{array}\right.=>\left\{\begin{array}{c}x=36\\y=25\end{array}\right.$