Question

Решите пожалуйста 30 баллов

Answer 1

$\left\{\begin{matrix}\frac{x^2+y^2}{xy}=2\\ \frac{x^3-y^2}{xy}=1\end{matrix}\right.\overset{x=y\neq 0}{\Leftrightarrow }\left\{\begin{matrix}x^2-2xy+y^2=0\\ x^3-xy-y^2=0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}\left ( x-y \right )^2=0\\ x^3-xy-y^2=0\end{matrix}\right.\\x=y\Rightarrow y^3-2y=0\Rightarrow y=2$

Answer 2

Ответ:

$\[(x;y) = (2;2)\]$

Объяснение:

$\[\begin{array}{l}\left\{ \begin{array}{l}\frac{{{x^2} + {y^2}}}{{xy}} = 2\\\frac{{{x^3} - {y^2}}}{{xy}} = 1\end{array} \right.{b^2} = 4a`\\xy = a\\x - y = b\\(1)\,\frac{{{b^2} + 2a}}{a} = 2 = > \,{b^2} + 2a = 2a = > \,{b^2} = 0 = > {(x - y)^2} = 0 = > x - y = 0 = > x = y\\(2)\frac{{{x^3} - {x^2}}}{{{x^2}}} = 1 = > \frac{{{x^2}(x - 1)}}{{{x^2}}} = 1 = > x - 1 = 1 = > x = 2 = > y = 2\\(x;y) = (2;2)\end{array}\]$