Huhu giúp e câu này với aa,e Camon aa

Giải thích các bước giải:

a.Để $A$ xác định

$\to\begin{cases}x\ne 0\\8-4x+2x^2-x^3\ne0\end{cases}$

$\to\begin{cases}x\ne 0\\  -\left(x-2\right)\left(x^2+4\right)\ne0\end{cases}$

$\to\begin{cases}x\ne 0\\  x\ne -2\end{cases}$

Ta có:

$A=(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{8-4x+2x^2-x^3})\cdot (1-\dfrac1x-\dfrac2{x^2})$

$\to A=(\dfrac{x^2-2x}{2x^2+8}-\dfrac{2x^2}{-(x-2)(x^2+4)})\cdot \dfrac{x^2-x-2}{x^2}$

$\to A=(\dfrac{x^2-2x}{2(x^2+4)}+\dfrac{2x^2}{(x-2)(x^2+4)})\cdot \dfrac{x^2-x-2}{x^2}$

$\to A=\dfrac{\left(x^2-2x\right)\left(x-2\right)+4x^2}{2\left(x-2\right)\left(x^2+4\right)}\cdot \dfrac{x^2-x-2}{x^2}$

$\to A=\dfrac{x^3+4x}{2\left(x-2\right)\left(x^2+4\right)}\cdot \dfrac{(x-2)(x+1)}{x^2}$

$\to A=\dfrac{x}{2\left(x-2\right)}\cdot \dfrac{(x-2)(x+1)}{x^2}$

$\to A=\dfrac{x+1}{2x}$

b.Để $A\in Z$

$\to x+1\quad\vdots\quad 2x$

$\to 2x+2\quad\vdots\quad 2x$

$\to 2\quad\vdots\quad 2x$

$\to 2x\in U(2)$

Mà $2x$ chẵn

$\to 2x\in\{2, -2\}$

$\to x\in\{1, -1\}$