Đáp án:
$a) -\dfrac{1}{x-2}$
b) $\left[ \begin{array}{l}x=\dfrac{2}{3}\\x=\dfrac{2}{5}\end{array} \right.$
c) $x\in \left \{ 2;+\infty \right \}$$
Giải thích các bước giải:
$a) \left ( \dfrac{x}{x^{2}-4}+\dfrac{2}{2-x}+\dfrac{1}{x+2} \right ):\left ( x-2+\dfrac{10-x^{2}}{x+2} \right )$
$=\left ( \dfrac{x}{(x-2).(x+2)}+\dfrac{2}{-(x-2)}+\dfrac{1}{x+2} \right ):\left ( x-2+\dfrac{10-x^{2}}{x+2} \right )$
$=\left ( \dfrac{x}{(x-2)(x+2)}-\dfrac{2}{x-2}+\dfrac{1}{x+2} \right ):\left ( x-2+\dfrac{10-x^{2}}{x+2} \right )$
$=\dfrac{x-2(x+2)+x-2}{(x-2)(x+2)}:\left ( x-2+\dfrac{10-x^{2}}{x+2} \right )$
$=\dfrac{x-2x-4+x-2}{(x-2)(x+2)}:\left ( x-2+\dfrac{10-x^{2}}{x+2} \right )$
$=\dfrac{0-6}{(x-2)(x+2)}:\left ( x-2+\dfrac{10-x^{2}}{x+2} \right )$
$=\dfrac{-6}{(x-2)(x+2)}:\left ( x-2+\dfrac{10-x^{2}}{x+2} \right )$
$=-\dfrac{6}{(x-2)(x+2)}:\dfrac{x(x+2)-2(x+2)+10-x^{2}}{x+2}$
$=-\dfrac{6}{(x-2)(x+2)}:\dfrac{x^{2}+2x-2x-4+10-x^{2}}{x+2}$
$=-\dfrac{6}{(x-2)(x+2)}:\dfrac{6}{x+2}$
$=-\dfrac{6}{(x-2)(x+2)}.\frac{x+2}{6}$
$=-\dfrac{1}{x-2}$
b) $|x|=\dfrac{1}{2}\Rightarrow \left[ \begin{array}{l}x=\dfrac{1}{2}\\x=-\dfrac{1}{2}\end{array} \right.$
* Với $x=\dfrac{1}{2}$
$\Rightarrow -\dfrac{1}{\frac{1}{2}-2}=-\dfrac{1}{-\dfrac{3}{2}}=\dfrac{1}{\dfrac{3}{2}}=\dfrac{2}{3}$
* Với $x=-\dfrac{1}{2}$
$\Rightarrow -\dfrac{1}{-\dfrac{1}{2}-2}=-\dfrac{1}{-\dfrac{5}{2}}=\dfrac{1}{\dfrac{5}{2}}=\dfrac{2}{5}$
c) $-\dfrac{1}{x-2}<0,x\neq 0$
$\Leftrightarrow \dfrac{1}{x-2}>0$
$\Leftrightarrow x-2>0$
$\Leftrightarrow x>2,x\neq 2$
$\Leftrightarrow x\in \left \{ 2;+\infty \right \}$
