$\frac{x-2}{2+x}$ - $\frac{3}{x-2}$ = $\frac{2(x-11)}{x^{2}-4}$
⇔ $\frac{(x-2)^{2} - 3(2+x)}{(2+x)(x-2)}$ = $\frac{2x-22}{(x-2)(x+2)}$
⇔ $\frac{x^{2} - 4x+4 - 6-3x)}{(2+x)(x-2)}$ = $\frac{2x-22}{(x-2)(x+2)}$
⇔ $\frac{x^{2}-7x-2}{(2+x)(x-2)}$ - $\frac{2x-22}{(x-2)(x+2)}$ = 0
⇔ $\frac{x^{2}-7x-2}{(2+x)(x-2)}$ - $\frac{2x-22}{(x-2)(x+2)}$ = 0
⇔ $\frac{x^{2}-7x-2-(2x-22)}{(2+x)(x-2)}$ = 0
⇔ $\frac{x^{2}-9x+20}{(2+x)(x-2)}$ = 0
⇒ $x^{2}$-9x+20 = 0
⇔ $x^{2}$-4x -5x +20 = 0
⇔ x(x-4) -5(x - 4)= 0
⇔ (x-4)(x-5)= 0
⇔\(\left[ \begin{array}{l}x-4=0\\x-5=0\end{array} \right.\)
⇔\(\left[ \begin{array}{l}x=4\\x=5\end{array} \right.\)
Vậy.......
