$\frac{x+8}{x^2+x-6}$ + $\frac{x-8}{x^2-x-2}$ = $\frac{2x+8}{x^2+x-6}$

Đáp án + Giải thích các bước giải:

$ĐKXĐ: x \ne -1; x \ne 2; x \ne -3$

 $\dfrac{x + 8}{x^2 + x - 6} + \dfrac{x - 8}{x^2 - x - 2} = \dfrac{2x + 8}{x^2 + x - 6}$

$\Leftrightarrow \dfrac{x - 8}{x^2 - x - 2} = \dfrac{2x + 8}{x^2 + x - 6} - \dfrac{x + 8}{x^2 + x - 6}$

$\Leftrightarrow \dfrac{x - 8}{x^2 - x - 2} = \dfrac{2x + 8 - x - 8}{x^2 + x - 6}$

$\Leftrightarrow \dfrac{x - 8}{x^2 - x - 2} = \dfrac{x}{x^2 + x - 6}$

$\Leftrightarrow \dfrac{x - 8}{(x + 1)(x - 2)} = \dfrac{x}{(x - 2)(x + 3)}$

$\Leftrightarrow \dfrac{(x - 8)(x + 3)}{(x + 1)(x - 2)(x + 3)} = \dfrac{x(x + 1)}{(x + 1)(x - 2)(x + 3)}$

$\Leftrightarrow \dfrac{x^2 - 5x - 24}{(x + 1)(x - 2)(x + 3)} = \dfrac{x^2 + x}{(x + 1)(x - 2)(x + 3)}$

$\Leftrightarrow x^2 - 5x - 24 = x^2 + x$

$\Leftrightarrow x^2 - 5x - 24 - x^2 - x = 0$

$\Leftrightarrow -6x - 24 = 0$

$\Leftrightarrow -6x = 24$

$\Leftrightarrow x = -4(tm)$

Vậy $S = $`{-4}`