Question

Дробь в ответе должна быть сокращённой. Быстро подалуйста ​

Answer 1

$\displaystyle\bf\\\left \{ {{\dfrac{1}{x+y} +\dfrac{1}{x-y} =17} \atop {\dfrac{3}{x+y} +\dfrac{9}{x-y}=105 }} \right. \\\\\\\frac{1}{x+y} =a \ \ ; \ \ \frac{1}{x-y} =b\\\\\\\left \{ {{a+b=17} \atop {3a+9b=105}} \right. \\\\\\-\left \{ {{a+b=17} \atop {a+3b=35}} \right. \\--------\\-2b=-18\\\\b=9\\\\a=17-b=17-9=8\\\\\\\left \{ {{\dfrac{1}{x+y} =8} \atop {\dfrac{1}{x-y}=9 }} \right. \\\\\\+\left \{ {{x+y=\dfrac{1}{8} } \atop {x-y=\dfrac{1}{9} }} \right. \\--------$

$\displaystyle\bf\\2x=\frac{17}{72} \\\\\\x=\frac{17}{144}\\\\\\y=\frac{1}{8} -x=\frac{1}{8} -\frac{17}{144} =\frac{18}{144} -\frac{17}{144} =\frac{1}{144}\\\\\\Otvet:\left \{ {{x=\dfrac{17}{144} } \atop {y=\dfrac{1}{144} }} \right.$

Answer 2

$\displaystyle \left \{ {{\frac{1}{x+y}+\frac{1}{x-y}=17 } \atop {\frac{3}{x+y}+\frac{9}{x-y}=105 }} \right. \\\\ \left \{ {{t+u=17} \atop {3t+9u=105}} \right. \\\\ (t,u)=(8,9)\\\\ \left \{ {{\frac{1}{x+y}=8 } \atop {\frac{1}{x-y}=9 }} \right.\\\\ (x,y)=(\frac{17}{144},\frac{1}{144})$