Đáp án:
$\dfrac{x-2}{x+2} - \dfrac{3}{x-2} =\dfrac{2(x-11)}{x^2-4}$
$\text{ĐKXĐ : $x \neq ± 2 $ }$
$⇔ \dfrac{(x-2)(x-2)}{(x+2)(x-2)} - \dfrac{3(x+2)}{(x-2)(x+2)} = \dfrac{2(x-11)}{(x-2)(x+2)}$
$⇒(x-2)^2 - 3(x+2) = 2(x-11)$
$⇔x^2 -4x +4 -3x -6 = 2x -22$
$⇔x^2 -4x -3x -2x +4 -6 +22 =0$
$⇔x^2 -9x +20 =0$
$⇔x^2 -2 . x . \dfrac{9}{2} + (\dfrac{9}{2})^2 - \dfrac{1}{4} = 0$
$⇔ (x - \dfrac{9}{2})^2 - \dfrac{1}{4} =0$
$⇔ (x-\dfrac{9}{2})^2 - (\dfrac{1}{2})^2 =0$
$⇔ (x - \dfrac{9}{2} - \dfrac{1}{2} ).( x - \dfrac{9}{2} + \dfrac{1}{2}) =0$
$⇔ $\(\left[ \begin{array}{l}x-\dfrac{9}{2} - \dfrac{1}{2}=0\\x-\dfrac{9}{2} +\dfrac{1}{2}=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}x=5 (TM)\\x=4 (TM)\end{array} \right.\)
$\text{Vậy phương trình có tập nghiệm S ={5 ; 4 } }$
