Đáp án:
$S=\bigg\{\dfrac{5+\sqrt{73}}{2};\dfrac{5-\sqrt{73}}{2}\bigg\}$
Giải thích các bước giải:
$\dfrac{x-2}{x+2}-\dfrac{2x-10}{4-x^2}=\dfrac{3}{x-2}$ ĐK: $x\neq\pm2$
$⇔\dfrac{x-2}{x+2}+\dfrac{2x-10}{x^2-4}=\dfrac{3}{x-2}$
$⇔\dfrac{(x-2)^2}{(x-2)(x+2)}+\dfrac{2x-10}{(x-2)(x+2)}=\dfrac{3(x+2)}{(x-2)(x+2)}$
$⇒x^2-4x+4+2x-10=3x+6$
$⇔x^2-4x+4+2x-10-3x-6=0$
$⇔x^2-5x-12=0$
\(⇔\left[ \begin{array}{l}x=\dfrac{5+\sqrt{73}}{2}(tm)\\x=\dfrac{5-\sqrt{73}}{2}(tm)\end{array} \right.\)
Vậy $S=\bigg\{\dfrac{5+\sqrt{73}}{2};\dfrac{5-\sqrt{73}}{2}\bigg\}$
