Giải thích các bước giải:

Bài 3:

a.Để A xác định

$\to\begin{cases}x^2+2x+1\ne 0\\ x^2-1\ne 0\\ x^3+x^2-x-1\ne 0\\ 2x^2+x\ne 0\end{cases}$ 

$\to\begin{cases}x\ne -1\\ x\ne \pm 1\\x\ne \pm 1 \\x\ne 0,-\dfrac 12\end{cases}$ 

$\to x\notin\{1,-1,0,-\dfrac 12\}$

Ta có :
$A=(\dfrac{x+2}{x^2+2x+1}-\dfrac{x-2}{x^2-1}):\dfrac{2x^2+x}{x^3+x^2-x-1}$

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

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

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

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

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

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

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

b.Ta có :
$x=-3\to A=\dfrac{2}{2(-3)+1}=\dfrac{-2}5$

$x=\dfrac{1}{4}\to A=\dfrac{2}{2(\dfrac{1}{4})+1}=\dfrac{4}3$

$x=\dfrac{-1}{2}\to $không xác định được A vì $x\ne -\dfrac 12$

c.Để $A=3\to \dfrac{2}{2x+1}=3\to 2x+1=\dfrac{3}2\to x=\dfrac{1}{4}$