Question

с подробным решением​

Answer 1

Ответ:

$\displaystyle 1)\ (-9;-7),\ (9;7)\\\\2)\ (1;2)$

Объяснение:

1)

$\displaystyle x\neq y\\\\\begin{cases}\frac{x^2+y^2}{10}=13\ \ \ |\cdot 10\\ \frac{x+y}{x-y}=8\ \ \ |\cdot(x-y)\end{cases}\\\\\\\begin{cases}x^2+y^2=130\\ x+y=8(x-y)\end{cases}\\\\\\\begin{cases}x^2+y^2=130\\ x+y=8x-8y\end{cases}\\\\\\\begin{cases}x^2+y^2=130\\ x-8x=-8y-y\end{cases}\\\\\\\begin{cases}x^2+y^2=130\\ -7x=-9y\ \ \ |:(-7)\end{cases}\\\\\\\begin{cases}x^2+y^2=130\\ x=\frac{9}{7}y\end{cases}$

$\displaystyle \begin{cases}(\frac{9}{7}y)^2+y^2=130\\ x=\frac{9}{7}y\end{cases}\\\\\\\begin{cases}\frac{81}{49}y^2+y^2=130\\ x=\frac{9}{7}y\end{cases}\\\\\\\begin{cases}\frac{130}{49}y^2=130\ \ \ |:\frac{130}{49}\\ x=-\frac{9}{7}y\end{cases}\\\\\\\begin{cases}y^2=49\\ x=-\frac{9}{7}y\end{cases}$

$\displaystyle \begin{cases}y=-7\\ x=-\frac{9}{7}y\end{cases}\ \ \ \ \ \ \ \ \ \ \begin{cases}y=7\\ x=\frac{9}{7}y\end{cases}\\\\\\\begin{cases}y=-7\\ x=-\frac{9}{7}y\end{cases}\ \ \ \ \ \ \ \ \ \ \begin{cases}y=7\\ x=\frac{9}{7}y\end{cases}\\\\\\\begin{cases}y=-7\\ x=\frac{9}{7}\cdot(-7)\end{cases}\ \ \ \ \ \ \begin{cases}y=7\\ x=\frac{9}{7}\cdot 7\end{cases}\\\\\\\begin{cases}y=-7\\ x=-9\end{cases}\ \ \ \ \ \ \ \ \ \ \ \ \ \begin{cases}y=7\\ x=9\end{cases}\\\\\\(-9;-7),\ (9;7)$

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2)

$\displaystyle y\neq 0,\ \ x\neq -2y\\\\\begin{cases}\frac{5}{x+2y}+\frac{8}{y}=5\ \ \ |\cdot (-2)\\ \frac{10}{x+2y}-\frac{2}{y}=1\end{cases}\\\\\\\begin{cases}-\frac{10}{x+2y}-\frac{16}{y}=-10\\ \frac{10}{x+2y}-\frac{2}{y}=1\end{cases}$

+__________________

$\displaystyle-\frac{16}{y}-\frac{2}{y}=-9\\\\-\frac{18}{y}=-9\\\\y=2\\\\\\\frac{10}{x+2y}-\frac{2}{y}=1\\\\\frac{10}{x+2\cdot 2}-\frac{2}{2}=1\\\\\frac{10}{x+4}-1=1\\\\\frac{10}{x+4}=1+1\\\\\frac{10}{x+4}=2\\\\x+4=\frac{10}{2}\\\\x+4=5\\\\x=5-4\\\\x=1\\\\(1;2)$