Question

Помогите умоляю !!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Answer 1

$1)\ \ \left\{\begin{array}{l}\dfrac{1}{x+y}+\dfrac{1}{x-y}=16\\\dfrac{2}{x+y}+\dfrac{9}{x-y}=95\end{array}\right\ \ \left\{\begin{array}{l}t=\dfrac{1}{x+y}\\p=\dfrac{1}{x-y} \end{array}\right\ \ \left\{\begin{array}{l}t+p=16\\2t+9p=95\end{array}\right$

$\left\{\begin{array}{l}2t+2p=32\\2t+9p=95\end{array}\right\ \ominus \ \left\{\begin{array}{l}-7p=-63\\t+p=16\end{array}\right\ \ \left\{\begin{array}{l}p=9\\t=16-p\end{array}\right\ \ \left\{\begin{array}{l}p=9\\t=7\end{array}\right$

$\left\{\begin{array}{l}\dfrac{1}{x+y}=7\\\dfrac{1}{x-y}=9\end{array}\right\ \ \left\{\begin{array}{l}x+y=\dfrac{1}{7}\\x-y=\dfrac{1}{9}\end{array}\right\ \oplus \ominus \ \left\{\begin{array}{l}2x=\dfrac{1}{7}+\dfrac{1}{9}\\\ 2y=\dfrac{1}{7}-\dfrac{1}{9}\end{array}\right\ \ \left\{\begin{array}{l}2x=\dfrac{16}{63}\\\ \ 2y=\dfrac{2}{63} \end{array}\right$

$\left\{\begin{array}{l}x=\dfrac{8}{63}\\\ \ y=\dfrac{1}{63}\end{array}\right\ \ \ \ \Rightarrow \ \ \ Otvet:\ \ \Big(\ \dfrac{8}{63}\ ;\ \dfrac{1}{63}\ \Big)\ .$

$2)\ \ \left\{\begin{array}{l}(5x+y)(x+2y)=\dfrac{266}{81}\\\dfrac{5x+y}{x+2y}=\dfrac{14}{19}\end{array}\right\ \ \left\{\begin{array}{l}t=5x+y\\p=x+2y\end{array}\right\ \ \left\{\begin{array}{l}t\cdot p=\dfrac{266}{81}\\\dfrac{t}{p}=\dfrac{14}{19}\end{array}\right\ \ \left\{\begin{array}{l}t=\dfrac{266}{81\cdot p}\\\dfrac{266}{81\cdot p^2}=\dfrac{14}{19}\end{array}\right$

$\left\{\begin{array}{l}t=\dfrac{266}{81\, p}\\p^2=\dfrac{266\cdot 19}{81\cdot 14} \end{array}\right\ \ \left\{\begin{array}{l}t=\dfrac{266}{81p}\\p^2=\dfrac{2\cdot 19\cdot 7\cdot 19}{81\cdot 2\cdot 7}=\dfrac{19^2}{9^2}\end{array}\right\ \ \left\{\begin{array}{l}\ t_1=-\dfrac{266\cdot 9}{81\cdot 19}=-\dfrac{14}{9}\ ,\ t_2=\dfrac{14}{9}\\p_1=-\dfrac{19}{9}\ ,\ p_2=\dfrac{19}{9}\end{array}\right$

$a)\ \left\{\begin{array}{l}5x+y=-\dfrac{14}{9}\\x+2y=\ -\dfrac{19}{9}\end{array}\right\ \ \left\{\begin{array}{l}45x+9y=-14\\9x+18y=-19\, |\cdot {-5}\end{array}\right\ \oplus \left\{\begin{array}{l}-81y=81\\x=-\dfrac{19}{9}-2y\end{array}\right$

$\left\{\begin{array}{l}y=-1\\x=-\dfrac{19}{9}+2\end{array}\right\ \ \ \left\{\begin{array}{l}y=-1\\x=-\dfrac{1}{9}\end{array}\right\ \ \ \to \ \ \ \Big(-\dfrac{1}{9}\ ;\ -1\Big)$

$b)\ \left\{\begin{array}{l}\ 5x+y=\dfrac{14}{9}\\x+2y=\dfrac{19}{9}\end{array}\right\ \ \left\{\begin{array}{l}45x+9y=14\\9x+18y=19\, |\cdot {-5}\end{array}\right\ \oplus \left\{\begin{array}{l}-81y=-81\\x=\dfrac{19}{9}-2y\end{array}\right$

$\left\{\begin{array}{l}y=1\\x=\dfrac{19}{9}-2\end{array}\right\ \ \ \left\{\begin{array}{l}y=1\\x=\dfrac{1}{9}\end{array}\right\ \ \ \to \ \ \ \Big(\ \dfrac{1}{9}\ ;\ 1\Big)\\\\\\Otvet:\ \ \Big(-\dfrac{1}{9}\ ;\ -1\Big)\ ,\ \Big(\ \dfrac{1}{9}\ ;\ 1\Big)\ .$