Đáp án:

\(\begin{array}{l}
a)\dfrac{5}{3}\\
b)\dfrac{1}{{x + 2}}\\
c)x =  - 1
\end{array}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)Thay:x = 3\\
 \to A = \dfrac{{{3^2} - 4}}{3} = \dfrac{5}{3}\\
b)B = \dfrac{5}{{x + 2}} - \dfrac{{2x - 12}}{{{x^2} - 4}} + \dfrac{2}{{2 - x}}\\
 = \dfrac{{5\left( {x - 2} \right) - 2x + 12 - 2\left( {x + 2} \right)}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
 = \dfrac{{5x - 10 - 2x + 12 - 2x - 4}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
 = \dfrac{{x - 2}}{{\left( {x + 2} \right)\left( {x - 2} \right)}}\\
 = \dfrac{1}{{x + 2}}\\
c)S = A.B = \dfrac{{{x^2} - 4}}{x}.\dfrac{1}{{x + 2}}\\
 = \dfrac{{\left( {x - 2} \right)\left( {x + 2} \right)}}{{x\left( {x + 2} \right)}}\\
 = \dfrac{{x - 2}}{x} = 1 - \dfrac{2}{x}\\
S \in Z \to \dfrac{2}{x} \in Z\\
 \to x \in U\left( 2 \right)\\
 \to \left[ \begin{array}{l}
x = 2\left( l \right)\\
x =  - 2\left( l \right)\\
x = 1\\
x =  - 1
\end{array} \right. \to \left[ \begin{array}{l}
S = 1 - \dfrac{2}{1} =  - 1\left( {KTM} \right)\\
S = 1 - \dfrac{2}{{ - 1}} = 3\left( {TM} \right)
\end{array} \right.\\
 \to x =  - 1
\end{array}\)