Đáp án:
$\begin{array}{l}
a)M = \left( {\frac{{2x}}{{x + 3}} + \frac{x}{{x - 3}} - \frac{{3{x^2} + 3}}{{{x^2} - 9}}} \right)\left( {x \ne 3;x \ne - 3} \right)\\
M = \frac{{2x}}{{x + 3}} + \frac{x}{{x - 3}} - \frac{{3{x^2} + 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \frac{{2x\left( {x - 3} \right) + x\left( {x + 3} \right) - 3{x^2} - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \frac{{2{x^2} - 6x + {x^2} + 3x - 3{x^2} - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}}\\
= \frac{{ - 3x - 3}}{{\left( {x - 3} \right)\left( {x + 3} \right)}} = \frac{{ - 3\left( {x + 1} \right)}}{{{x^2} - 9}}\\
b)\left| {x - 1} \right| = 3\\
\Rightarrow \left[ \begin{array}{l}
x - 1 = 3\\
x - 1 = - 3
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 4\\
x = - 2
\end{array} \right.\\
+ Khi\,x = 4 \Rightarrow M = \frac{{ - 3.5}}{{{4^2} - 9}} = - \frac{{15}}{7}\\
+ Khi\,x = - 2 \Rightarrow M = \frac{{ - 3.\left( { - 1} \right)}}{{4 - 9}} = \frac{3}{5}\\
c)M < 0\\
\Rightarrow \frac{{ - 3\left( {x + 1} \right)}}{{{x^2} - 9}} < 0\\
\Rightarrow \frac{{x + 1}}{{\left( {x + 3} \right)\left( {x - 3} \right)}} > 0\\
\Rightarrow \left[ \begin{array}{l}
- 3 < x < - 1\\
x > 3
\end{array} \right.
\end{array}$
