`color{#FF0000}{e}color{#FF7F00}{n}color{#FFFF00}{d}color{#00FF00}{f}color{#0000FF}{i}color{#4B0082}{e}color{#8A2BE2}{ld}`

a) ĐKXĐ: $x \neq \pm 1, x \neq 0, x \neq -2003$

Biến đổi biểu thức trong ngoặc:

$= \frac{(x+1)^2 - (x-1)^2 + x^2-4x-1}{(x-1)(x+1)}$

$= \frac{x^2+2x+1 - (x^2-2x+1) + x^2-4x-1}{x^2-1}$

$= \frac{x^2+2x+1 - x^2+2x-1 + x^2-4x-1}{x^2-1}$

$= \frac{x^2-1}{x^2-1} = 1$

$\Rightarrow K = 1 : \frac{x+2003}{x}$

$\Rightarrow K = \frac{x}{x+2003}$

b) Ta có: $K = \frac{x}{x+2003} = \frac{x+2003-2003}{x+2003} = 1 - \frac{2003}{x+2003}$

Để $K$ nguyên thì $2003 \vdots (x+2003)$

$\Rightarrow x+2003 \in Ư(2003) = \{1; -1; 2003; -2003\}$

$x+2003 = 1 \Rightarrow x = -2002$ (TM)

$x+2003 = -1 \Rightarrow x = -2004$ (TM)

$x+2003 = 2003 \Rightarrow x = 0$ (Loại vì ĐKXĐ)

$x+2003 = -2003 \Rightarrow x = -4006$ (TM)

Vậy $x \in \{-2002; -2004; -4006\}$