[tex]\dfrac{3}{(2-x)^2} -\dfrac{5}{(x+2)^2}=\dfrac{14}{x^2-4} \\ \\ \dfrac{3}{\Big[-(x-2)\Big]^2} - \dfrac{5}{(x+2)^2} = \dfrac{14}{(x-2)(x+2)} \\ \\ \dfrac{3}{(x-2)^2} - \dfrac{5}{(x+2)^2} = \dfrac{14}{(x-2)(x+2)} \Big|\cdot \underset{numitorul~comun}{\underbrace{(x-2)^2\cdot (x+2)^2}}} \\ \\ 3(x+2)^2 - 5(x-2)^2 = 14(x-2)(x+2)\\ \\3(x^2+4x+4) - 5(x^2-4x+4) = 14(x^2-4) \\ \\ 3x^2+12x+12-5x^2+20x-20 = 14x^2-56 \\ \\ (3x^2-5x^2-14x^2) +(12x+20x) + (12-20+56) = 0 \\ \\ -16x^2 + 32x +48 = 0 \Big|:(-16) [/tex]
[tex]x^2 - 2x - 3 = 0 \\ x^2+x-3x-3 = 0 \\ x(x+1)-3(x+1) = 0 \\ (x+1)(x-3) = 0 \\ \\ x_1 = -1,\quad x_2 = 3 \\ \\ D = \mathbb{R} \backslash \Big\{-2,2\Big\} \\ \\ \Rightarrow \boxed{S = \Big\{-1,3\Big\}}[/tex]
