Giải thích các bước giải:
a.Ta có:
$A=(\dfrac{x}{x^2-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2}):(x-2+\dfrac{10-x^2}{x+2})$
$\to A=(\dfrac{x}{(x-2)(x+2)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}):\dfrac{(x-2)(x+2)+10-x^2}{x+2}$
$\to A=(\dfrac{x}{(x-2)(x+2)}-\dfrac{2(x+2)}{(x-2)(x+2)}+\dfrac{x-2}{(x+2)(x-2)}):\dfrac{x^2-4+10-x^2}{x+2}$
$\to A=\dfrac{x-2(x+2)+x-2}{(x+2)(x-2)}:\dfrac{6}{x+2}$
$\to A=\dfrac{-6}{(x+2)(x-2)}\cdot \dfrac{x+2}{6}$
$\to A=\dfrac{-1}{x-2}$
b.Ta có $|x|=\dfrac12\to x=\dfrac12$ hoặc $x=-\dfrac12$
Nếu $x=\dfrac12\to A=\dfrac{-1}{\dfrac12-2}=\dfrac23$
$x=-\dfrac12\to A=\dfrac{-1}{-\dfrac12-2}=\dfrac25$
c.Để $A<0$
$\to \dfrac{-1}{x-2}<0$
$\to x-2>0$ vì $-1<0$
$\to x>2$
