\(\begin{array}{l}
A = \left( {\dfrac{{ - 3}}{{x - 2}} + \dfrac{1}{{x + 2}}} \right):\dfrac{x}{{x + 2}}\\
a)\,\,DKXD:\,\,\left\{ \begin{array}{l}
x - 2 \ne 0\\
x + 2 \ne 0
\end{array} \right. \Leftrightarrow \,\left\{ \begin{array}{l}
x \ne 2\\
x \ne - 2
\end{array} \right.\\
b)\,\,A = \left( {\dfrac{{ - 3}}{{x - 2}} + \dfrac{1}{{x + 2}}} \right):\dfrac{x}{{x + 2}}\\
= \dfrac{{ - 3\left( {x + 2} \right) + 1.\left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}:\dfrac{x}{{x + 2}}\\
= \dfrac{{ - 3x - 6 + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{x} = \dfrac{{ - 2x - 8}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}.\dfrac{{x + 2}}{x}\\
= \dfrac{{ - 2\left( {x + 4} \right)\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)x}} = \dfrac{{ - 2\left( {x + 4} \right)}}{{x\left( {x - 2} \right)}}\\
c)\,\,Voi\,\,x = 5\,\,ta\,\,co:\\
A = \dfrac{{ - 2.\left( {5 + 4} \right)}}{{5.\left( {5 - 2} \right)}} = \dfrac{{ - 18}}{{15}} = \dfrac{{ - 6}}{5}
\end{array}\)
