Giải thích các bước giải:
a.Khi $x=-3$
$\to A=\dfrac{-3+1}{-3+2}=\dfrac{-2}{-1}=2$
b.Ta có:
$B=\dfrac{3}{x-1}-\dfrac{x+5}{x^2-1}$
$\to B=\dfrac{3}{x-1}-\dfrac{x+5}{(x-1)(x+1)}$
$\to B=\dfrac{3(x+1)}{(x-1)(x-1)}-\dfrac{x+5}{(x-1)(x+1)}$
$\to B=\dfrac{3(x+1)-(x+5)}{(x-1)(x+1)}$
$\to B=\dfrac{3x+3-x-5}{(x-1)(x+1)}$
$\to B=\dfrac{2x-2}{(x-1)(x+1)}$
$\to B=\dfrac{2(x-1)}{(x-1)(x+1)}$
$\to B=\dfrac{2}{x+1}$
c.Ta có:
$C=\dfrac{2A}{B}$
$\to C=\dfrac{2\cdot \dfrac{x+1}{x+2}}{\dfrac{2}{x+1}}$
$\to C=\dfrac{(x+1)^2}{x+2}$
Để $C\in Z$ vì $x\in Z$
$\to (x+1)^2\quad\vdots\quad x+2$
$\to (x+1)^2-1+1\quad\vdots\quad x+2$
$\to (x+1-1)(x+1+1)+1\quad\vdots\quad x+2$
$\to x(x+2)+1\quad\vdots\quad x+2$
$\to 1\quad\vdots\quad x+2$
$\to x+2\in U(1)$
$\to x+2\in \{1, -1\}$
$\to x\in\{-1 , -3\}$
