Đáp án:

Giải thích các bước giải:

ĐKXĐ: $x \neq \pm 3$

1) 

$x = -2$ (TM ĐKXĐ)

$A = \dfrac{2}{-2 - 3} = -\dfrac{2}{5}$

2) 

$B = \dfrac{x}{x + 3} + \dfrac{2x}{3 - x} + \dfrac{2x^2 + 12x}{x^2 - 9}$

$= \dfrac{x(x - 3) - 2x(x + 3) + 2x^2 + 12x}{(x - 3)(x + 3)}$

$= \dfrac{x^2 - 3x - 2x^2 - 6x + 2x^2 + 12x}{(x - 3)(x + 3)}$

$= \dfrac{x^2 + 3x}{(x - 3)(x + 3)} = \dfrac{x(x + 3)}{(x - 3)(x + 3)}$

$= \dfrac{x}{x - 3}$ (đpcm)

3) 

$B = 2x \Leftrightarrow \dfrac{x}{x - 3} = 2x$

$\Leftrightarrow x = 2x(x - 3) \Leftrightarrow 2x^2 - 7x = 0$

$\Leftrightarrow x(2x - 7) = 0 \Leftrightarrow \left[ \begin{array}{l} x = 0 \\ x = \dfrac{7}{2} \end{array} \right.$ (TM ĐKXĐ)

4)

$P = \dfrac{2}{x - 3} + \dfrac{x}{x - 3} = \dfrac{x + 2}{x - 3} = 1 + \dfrac{5}{x - 3}$

$P \in \mathbb{Z} \Leftrightarrow (x - 3) \in \text{Ư}(5) = \{ \pm 1; \pm 5 \}$

$\begin{array}{|c|c|c|c|c|} \hline x - 3 & -5 & -1 & 1 & 5 \\ \hline x & -2 & 2 & 4 & 8 \\ \hline  & \text{TM} & \text{TM} & \text{TM} & \text{TM} \\ \hline \end{array}$

Vậy $x \in \{ -2; 2; 4; 8 \}$