Đáp án:
$\text{$\dfrac{1}{3-x}$ + $\dfrac{14}{x^2-9}$ =1}$
$\text{ =>$\dfrac{-1}{x-3}$ + $\dfrac{14}{x^2-9}$ = 1}$
$\text{ĐKXĐ : x $\neq$ = 3}$
$\text{=> $\dfrac{-(x+3)}{(x-3)(x+3)}$ + $\dfrac{14}{x^2-9}$ = $\dfrac{(x^2-9}{x^2-9}$}$
$\text{=> -x-3 + 14 = x²-9}$
$\text{<=>-x² - x -3+14+9=0}$
$\text{<=>-x² - x + 20 =0}$
$\text{<=> (-x² + 4x) (-5x+20)=0}$
$\text{<=>-x(x-4)-5(x-4)=0}$
$\text{<=>(x-4)(-x-5)=0}$
$\text{<=>\(\left[ \begin{array}{l}x-4=0\\-x-5=0\end{array} \right.\) }$
$\text{<=>\(\left[ \begin{array}{l}x=4\\x=-5\end{array} \right.\) }$
$\text{Vậy phương trình có tập nghiệm S={4 ; -5}}$
