Giải thích các bước giải:
a.ĐKXĐ: $x\ne \pm2$
Ta có:
$P=(\dfrac{2}{x+2}-\dfrac{4}{x^2+4x+4}):(\dfrac{2}{x^2-4}+\dfrac1{2-x})$
$\to P=(\dfrac{2}{x+2}-\dfrac{4}{(x+2)^2}):(\dfrac{2}{(x-2)(x+2)}-\dfrac1{x-2})$
$\to P=(\dfrac{2(x+2)}{(x+2)^2}-\dfrac{4}{(x+2)^2}):(\dfrac{2}{(x-2)(x+2)}-\dfrac{x+2}{(x+2)(x-2)})$
$\to P=\dfrac{2(x+2)-4}{(x+2)^2}:\dfrac{2-(x+2)}{(x+2)(x-2)}$
$\to P=\dfrac{2x}{(x+2)^2}:\dfrac{-x}{(x+2)(x-2)}$
$\to P=\dfrac{2x}{(x+2)^2}\cdot\dfrac{(x+2)(x-2)}{-x}$
$\to P=-\dfrac{2(x-2)}{x+2}$
b.Để $P\in Z$
$\to -\dfrac{2(x-2)}{x+2}\in Z$
$\to \dfrac{2(x-2)}{x+2}\in Z$
Mà $x\in Z$
$\to 2(x-2)\quad\vdots\quad x+2$
$\to 2(x+2)-8\quad\vdots\quad x+2$
$\to 8\quad\vdots\quad x+2$
$\to x+2\in\{1, 2, 4, 8, -1, -2, -4, -8\}$
$\to x\in\{-1, 0, 2, 6, -3, -4, -6, -10\}$
Mà $x\ne \pm2$
$\to x\in\{-1, 0, 6, -3, -4, -6, -10\}$
c.Để $P\ge-\dfrac12$
$\to -\dfrac{2(x-2)}{x+2}\ge-\dfrac12$
$\to \dfrac{2(x-2)}{x+2}\le\dfrac12$
$\to \dfrac{2(x-2)}{x+2}-\dfrac12\le 0$
$\to \dfrac{3x-10}{x+2}\le 0$
$\to -2<x\le\dfrac{10}3$
