Пжжжжж!!!!!!!!!!!!!!!!!!!!. Created by lisenokty31. algebra-ru

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Verified answer

$\left\{\begin{array}{l}\dfrac{1}{x+y}+\dfrac{1}{x-y}=17\, |\cdot 6\\\dfrac{6}{x+y}+\dfrac{10}{x-y}=138\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{6}{x+y}+\dfrac{6}{x-y}=102\\\dfrac{6}{x+y}+\dfrac{6}{x-y}+\dfrac{4}{x-y}=138\end{array}\right\\\\\\\left\{\begin{array}{l}\dfrac{1}{x+y}+\dfrac{1}{x-y}=17\\102+\dfrac{4}{x-y}=138\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{1}{x+y}+\dfrac{1}{x-y}=17\\\dfrac{4}{x-y}=36\end{array}\right$

$\left\{\begin{array}{l}\dfrac{1}{x+y}+\dfrac{1}{x-y}=17\\x-y=\dfrac{1}{9}\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{1}{x+y}+\dfrac{1}{x-y}=17\\x=y+\dfrac{1}{9}\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{1}{2y+\frac{1}{9}}+\dfrac{1}{\frac{1}{9}}=17\\x=y+\dfrac{1}{9}\end{array}\right$

$\left\{\begin{array}{l}\dfrac{9}{18y+1}+9=17\\x=y+\dfrac{1}{9}\end{array}\right\ \ \left\{\begin{array}{l}\dfrac{9}{18y+1}=8\\x=y+\dfrac{1}{9}\end{array}\right\ \ \left\{\begin{array}{l}18y+1=\dfrac{9}{8}\\x=y+\dfrac{1}{9}\end{array}\right\ \ \left\{\begin{array}{l}18y=\dfrac{1}{8}\\x=y+\dfrac{1}{9}\end{array}\right$

$\left\{\begin{array}{l}y=\dfrac{1}{144}\\x=y+\dfrac{1}{9}\end{array}\right\ \ \left\{\begin{array}{l}y=\dfrac{1}{144}\\\, x=\dfrac{1}{144}+\dfrac{1}{9}\end{array}\right\ \ \left\{\begin{array}{l}y=\dfrac{1}{144}\\\, x=\dfrac{17}{144}\end{array}\right\\\\\\Otvet:\ \ \Big(\, \dfrac{17}{144}\ ;\ \dfrac{1}{144}\, \Big)\ .$