Question

Найдите корни уравнения
6/x^2-4-3/x-2=1/x+2

Answer 1

$frac{6}{x^2-4} - frac{3}{x-2}= frac{1}{x+2}; \ frac{6}{(x-2)(x+2)}- frac{3}{x-2}- frac{1}{x+2}=0 \ frac{6-3x-6-x+2}{(x-2)(x+2)}=0; \ frac{2-4x}{(x-2)(x+2)}=0; \ left { {{2-4x=0;} atop {x-2 neq 0;x+2 neq 0;}} right. \ left { {{-4x=-2;} atop {x neq 2;x neq -2;}} right. \ left { {{x= frac{1}{2} ;} atop {x neq 2;x neq -2;}} right.$