Question

Решите систему уравнений:
$frac{3}{2x+y} + frac{1}{2x-y} =frac{2}{5} \frac{7}{2x+y} + frac{2}{2x-y} =frac{3}{5}$

Answer 1

$left { {{}frac{3}{2x+y}+ frac{1}{2x-y}= frac{2}{5}  atop {frac{7}{2x+y}+ frac{2}{2x-y}= frac{3}{5}}} right. \ \ 2x+yneq 0;yneq -2x;\ \ 2x-yneq 0;yneq2x \ \ left { {{frac{-6}{2x+y}- frac{2}{2x-y}= frac{-4}{5}} atop {frac{7}{2x+y}+ frac{2}{2x-y}= frac{3}{5}}} right. \ \ ----------\ frac{-6+7}{2x+y} =frac{-1}{5} \ \ frac{1}{2x+y} =-frac{1}{5} \ \ 2x+y=-5\ \ y=-2x-5\ \ frac{3}{2x-2x-5}+ frac{1}{2x+2x+5}= frac{2}{5}\ \ frac{1}{4x+5} =frac{2}{5} +frac{3}{5}$

$frac{1}{4x+5} =1\ \ 4x+5=1\ \ 4x=-4\ \ x=-1\ \ y=-2*(-1)-5=-3\ \ OTVET:(-1;-3)$