Giải thích các bước giải:
a.Ta có:
$B=(\dfrac{x}{x^2-36}-\dfrac{x-6}{x^2+6x}):\dfrac{2x-6}{x^2+6x}$
$\to B=(\dfrac{x}{(x-6)(x+6)}-\dfrac{x-6}{x(x+6)})\cdot\dfrac{x^2+6x}{2x-6}$
$\to B=(\dfrac{x^2}{x(x-6)(x+6)}-\dfrac{(x-6)^2}{x(x+6)(x-6)})\cdot\dfrac{x(x+6)}{2x-6}$
$\to B=\dfrac{x^2-(x-6)^2}{x(x+6)(x-6)}\cdot\dfrac{x(x+6)}{2x-6}$
$\to B=\dfrac{12x-36}{x(x+6)(x-6)}\cdot\dfrac{x(x+6)}{2x-6}$
$\to B=\dfrac{6(2x-6)}{x(x+6)(x-6)}\cdot\dfrac{x(x+6)}{2x-6}$
$\to B=\dfrac{6}{x-6}$
b.Ta có: $x^2-2x=0\to x(x-2)=0\to x\in\{0,2\}$
Nếu $x=0\to B=\dfrac6{0-6}=-1$
Nếu $x=2\to B=\dfrac6{2-6}=\dfrac{-3}{2}$
c.Để $B=1$
$\to\dfrac6{x-6}=1\to x-6=6\to x=12$
d.Để $B\in Z$
$\to 6\quad\vdots\quad x-6$ vì $x\in Z$
$\to x-6\in\{1,2,3,6,-1,-2,-3,-6\}$
$\to x\in\{7,8,9,12, 5,4,3,0\}$
Mà $x\ne \pm6, 3$
$\to x\in\{7,8,9,12, 5,4,0\}$
