Giải thích các bước giải:

a.ĐKXĐ: $x\ne\pm3, x\ne -\dfrac12$

Ta có:

$P=(\dfrac{x-1}{x+3}+\dfrac2{x-3}+\dfrac{x^2+3}{9-x^2}):(\dfrac{2x-1}{2x+1}-1)$

$\to P=(\dfrac{x-1}{x+3}+\dfrac2{x-3}-\dfrac{x^2+3}{x^2-9}):\dfrac{-2}{2x+1}$

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

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

$\to P=\dfrac{-2x+6}{\left(x+3\right)\left(x-3\right)}\cdot \dfrac{2x+1}{-2}$

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

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

b.Ta có:

$|x+1|=\dfrac12\to x+1=\pm\dfrac12\to x\in\{-\dfrac32, -\dfrac12\}$

Mà $x\ne -\dfrac12\to x=-\dfrac32$

$\to P=\dfrac{3\cdot (-\dfrac32)+1}{-\dfrac32+3}$

$\to P=\dfrac{-7}3$

c.Ta có:

$P=\dfrac{x}2$

$\to \dfrac{2x+1}{x+3}=\dfrac{x}2$

$\to 4x+2=x^2+3x$

$\to x^2-x-2=0$

$\to (x-2)(x+1)=0$

$\to x\in\{2, -1\}$

d.Để $P\in Z$

$\to 2x+1\quad\vdots\quad x+3$

$\to 2(x+3)-5\quad\vdots\quad x+3$

$\to 5\quad\vdots\quad x+3$

$\to x+3\in U(5)$

$\to x+3\in\{1, 5, -1, -5\}$
$\to x\in\{-2, 2, -4, -8\}$