có x² - 4 = ( x -2 )(x + 2)
$\frac{x-2}{x + 2}$ - $\frac{3}{x -2 }$ = $\frac{2 (x - 11 )}{x^2 -4}$
⇔ $\frac{(x -2)(x-2)}{(x-2)(x+2)}$ - $\frac{3(x +2)}{(x +2)(x-2)}$ = $\frac{2(x-11)}{(x+2)( x- 2)}$
⇒ ( x-2)² - 3x + 6 = 2x - 22
⇔ x² - 4x + 4 - 3x - 6 = 2x - 22
⇔ x² -9x + 20 = 0
⇔ x² - 5x - 4x + 20 = 0
⇔ x( x - 5) - 4 (x - 5) = 0
⇔ (x-5)(x-4) = 0
⇔\(\left[ \begin{array}{l}x- 5 =0\\x- 4=0\end{array} \right.\)
⇔ \(\left[ \begin{array}{l}x=5\\x=4\end{array} \right.\)
vậy pt có 2 nghiệm S = { 5 ; 4}
