Đáp án:
= $\frac{1}{x-3}$
Giải thích các bước giải:
( $\frac{1}{x²-9}$ + $\frac{2}{3-x}$ + $\frac{3}{x+3}$ ) : $\frac{x-14}{x+3}$
= [ $\frac{1}{(x-3)(x+3)}$ - $\frac{2}{x-3}$ + $\frac{3}{x+3}$ ] : $\frac{x-14}{x+3}$
= [ $\frac{1}{(x-3)(x+3)}$ - $\frac{2(x+3)}{(x-3)(x+3)}$ + $\frac{3(x-3)}{(x+3)(x-3)}$ ] : $\frac{x-14}{x+3}$
= [$\frac{ 1 - 2(x+3) + 3(x-3)}{(x-3)(x+3)}$] : $\frac{x-14}{x+3}$
= [$\frac{1 - 2x - 6 + 3x - 9}{(x-3)(x+3)}$ ] : $\frac{x-14}{x+3}$
= $\frac{x-14}{(x-3)(x+3)}$ : $\frac{x-14}{x+3}$
= $\frac{x-14}{(x-3)(x+3)}$ . $\frac{x+3}{x-14}$
= $\frac{1}{x-3}$
