Giải thích các bước giải:
a.Ta có:
$A=(\dfrac{x^2}{x^3-4x}+\dfrac{6}{6-3x}+\dfrac{1}{x+2}):(x-2+\dfrac{10-x^2}{x+2})$
$\to A=(\dfrac{x^2}{x(x^2-4)}-\dfrac{6}{3x-6}+\dfrac{1}{x+2}):\dfrac{(x-2)(x+2)+10-x^2}{x+2}$
$\to A=(\dfrac{x^2}{x(x-2)(x+2))}-\dfrac{6}{3(x-2)}+\dfrac{1}{x+2}):\dfrac{x^2-4+10-x^2}{x+2}$
$\to A=(\dfrac{x^2}{x(x-2)(x+2)}-\dfrac{6(x+2)}{3(x+2)(x-2)}+\dfrac{x-2}{(x-2)(x+2)}):\dfrac{6}{x+2}$
$\to A=(\dfrac{x^2}{x(x-2)(x+2)}-\dfrac{2(x+2)}{(x+2)(x-2)}+\dfrac{x-2}{(x-2)(x+2)})\cdot \dfrac{x+2}{6}$
$\to A=(\dfrac{x}{(x-2)(x+2)}-\dfrac{2(x+2)}{(x+2)(x-2)}+\dfrac{x-2}{(x-2)(x+2)})\cdot \dfrac{x+2}{6}$
$\to A=(\dfrac{x-2(x+2)+x-2}{(x-2)(x+2)})\cdot \dfrac{x+2}{6}$
$\to A=(\dfrac{x-2x-4+x-2}{(x-2)(x+2)})\cdot \dfrac{x+2}{6}$
$\to A=(\dfrac{-6}{(x-2)(x+2)})\cdot \dfrac{x+2}{6}$
$\to A=\dfrac{-1}{x-2}$
$\to A=\dfrac{1}{2-x}$
b.Ta có $x=|0.5|\to x=0.5$ hoặc $x=-0.5$
$\to A=\dfrac1{2-0.5}=\dfrac23$ hoặc $A=\dfrac1{2+0.5}=\dfrac25$
c.Để $A=2$
$\to \dfrac{1}{2-x}=2\to 2-x=\dfrac12\to x=\dfrac32$
Để $A<0$
$\to \dfrac1{2-x}<0$
$\to 2-x<0$
$\to x>2$
d.Để $A\in Z$
$\to \dfrac{1}{2-x}\in Z$
$\to 1\quad\vdots\quad 2-x$ vì $x\in Z$
$\to 2-x\in\{1,-1\}$
$\to x\in\{1, 3\}$
