Question

Все на рисунке!!!!!!!!!!

Answer 1

Ответ:

(2,2)

Объяснение:

Рассмотрим первое уравнение:

$\frac{x^2 + y^2}{xy} = 2 \rightarrow x^2 + y^2 = 2xy \rightarrow x^2 - 2xy + y^2 = 0 \rightarrow (x-y)^2=0 \rightarrow x = y \ (x \neq 0, \ y \neq 0)$

Подставим:

$\frac{x^3 - y^2}{xy} = 1 \rightarrow \frac{x^3 - x^2}{x^2} = 1 \rightarrow x -1 = 1 \rightarrow x = 2 \rightarrow y = 2$

(2,2) - решение системы.

Answer 2

$\left\{\begin{array}{l}\dfrac{x^2+y^2}{xy}=2\\\dfrac{x^3-y^2}{xy}=1\end{array}\right\ \ x\ne 0\ ,\ y\ne 0\ \ \left\{\begin{array}{l}x^2+y^2=2xy\\x^3-y^2=xy\end{array}\right\ \oplus\ \left\{\begin{array}{l}x^2-2xy+y^2=0\\x^3+x^2=3xy\end{array}\right\\\\\\\left\{\begin{array}{l}(x-y)^2=0\\x\cdot (x^2+x-3y)=0\end{array}\right\ \ \left\{\begin{array}{l}x=y\ \ ,\ x\ne 0\ ,\ y\ne 0\\x^2+x-3y=0\end{array}\right$

$x^2+x-3x=0\ \ ,\ \ \ x^2-2x=0\ \ ,\ \ x\cdot (x-2)=0\ \ ,\ \ x=2\ \ (x\ne 0)\\\\y=x=2\\\\Otvet:\ \ (2;2)\ .$