$(\frac{1}{x-5}+$ $\frac{1}{x^{2}-25})$ :$\frac{x+6}{x^{2}-25}$ = 1

Đáp án + Giải thích các bước giải:

 $\bigg(\dfrac{1}{x - 5} + \dfrac{1}{x^2 - 25}\bigg) : \dfrac{x + 6}{x^2 - 25} = 1(x \ne \pm5, -6)$

$\Leftrightarrow \bigg(\dfrac{1}{x - 5} + \dfrac{1}{(x + 5)(x - 5)}\bigg) : \dfrac{x + 6}{(x + 5)(x - 5)} = 1$
$\Leftrightarrow \bigg(\dfrac{x + 5}{(x + 5)(x - 5)} + \dfrac{1}{(x + 5)(x - 5)}\bigg) : \dfrac{x + 6}{(x + 5)(x - 5)} = 1$

$\Leftrightarrow \dfrac{x + 5 + 1}{(x + 5)(x - 5)} : \dfrac{x + 6}{(x + 5)(x - 5)} = 1$
$\Leftrightarrow \dfrac{x + 6}{(x + 5)(x - 5)} : \dfrac{x + 6}{(x + 5)(x - 5)} = 1$(luôn đúng)

Vậy $S = $`{x | x \ne \pm 5, -6}`

$\color{red}{\text{#Vexi}}$