Đáp án:
$\begin{array}{l}
a)S = \left( {\dfrac{{3x + 6}}{{{x^2} - 4}} + \dfrac{x}{{x - 2}}} \right):\dfrac{{{x^2} - 9}}{{x - 3}}\\
= \left( {\dfrac{{3\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x - 2} \right)}} + \dfrac{x}{{x - 2}}} \right).\dfrac{{x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \left( {\dfrac{3}{{x - 2}} + \dfrac{x}{{x - 2}}} \right).\dfrac{1}{{x + 3}}\\
= \dfrac{{3 + x}}{{x - 2}}.\dfrac{1}{{x + 3}}\\
= \dfrac{1}{{x - 2}}\\
b)S = x - 2\\
\Leftrightarrow \dfrac{1}{{x - 2}} = x - 2\\
\Leftrightarrow {\left( {x - 2} \right)^2} = 1\\
\Leftrightarrow \left[ \begin{array}{l}
x - 2 = 1\\
x - 2 = - 1
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 3\left( {ktm} \right)\\
x = 1\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = 1\\
c)Q = 3 + S = 3 + \dfrac{1}{{x - 2}}\\
S \in Z\\
\Leftrightarrow \dfrac{1}{{x - 2}} \in Z\\
\Leftrightarrow x - 2 = - 1\left( {do:x - 2\# 1} \right)\\
\Leftrightarrow x = 1\\
Vậy\,x = 1
\end{array}$
