Lời giải:ĐK: \(\left\{\begin{array}{l} x-2 \neq 0 \\ x+2 \neq 0 \end{array} \Leftrightarrow\left\{\begin{array}{l} x \neq 2 \\ x \neq-2 \end{array}\right.\right.\)
Khi đó
\(\begin{aligned} &\text { g) } \frac{x-2}{x+2}+\frac{3}{2-x}=\frac{2(x-11)}{x^{2}-4}\\ &\Leftrightarrow \frac{x-2}{x+2}-\frac{3}{x-2}=\frac{2 x-22}{(x-2)(x+2)}\\ &\Leftrightarrow \frac{(x-2)(x-2)}{(x+2)(x-2)}-\frac{3(x+2)}{(x-2)(x+2)}=\frac{2 x-22}{(x-2)(x+2)}\\ &\Leftrightarrow(x-2)(x-2)-3(x+2)=2 x-22\\ &\Leftrightarrow(x-2)^{2}-3(x+2)-2 x+22=0\\ &\Leftrightarrow x^{2}-4 x+4-3 x-6-2 x+22=0\\ &\Leftrightarrow x^{2}-9 x+20=0\\ &\Leftrightarrow x^{2}-4 x-5 x+20=0\\ &\Leftrightarrow x(x-4)-5(x-4)=0\\ &\Leftrightarrow(x-4)(x-5)=0\\ &\Leftrightarrow\left[\begin{array}{l} x-4=0 \\ x-5=0 \end{array} \Leftrightarrow\left[\begin{array}{l} x=4(\mathrm{tm}) \\ x=5(t m) \end{array}\right.\right.\\ &\Rightarrow S=\{4 ; 5\} \end{aligned}\)
