Giải thích các bước giải:

a,

ĐKXĐ:  \(\left\{ \begin{array}{l}
{x^3} - 4x \ne 0\\
6 - 3x \ne 0\\
x + 2 \ne 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
x \ne 0\\
x \ne  \pm 2
\end{array} \right.\)

Ta có:

\(\begin{array}{l}
A = \frac{{{x^2}}}{{{x^3} - 4x}} + \frac{6}{{6 - 3x}} + \frac{1}{{x + 2}}\\
 = \frac{{{x^2}}}{{x\left( {{x^2} - 4} \right)}} + \frac{6}{{3\left( {2 - x} \right)}} + \frac{1}{{x + 2}}\\
 = \frac{x}{{\left( {x - 2} \right)\left( {x + 2} \right)}} - \frac{2}{{x - 2}} + \frac{1}{{x + 2}}\\
 = \frac{{x - 2\left( {x + 2} \right) + \left( {x - 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 = \frac{{x - 2x - 4 + x - 2}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
 = \frac{{ - 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
b,\\
M = A:\frac{{ - 3}}{{x + 2}} = \frac{{ - 6}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}:\frac{{ - 3}}{{\left( {x + 2} \right)}} = \frac{2}{{x - 2}}\\
c,\\
M = \frac{{ - 1}}{x}\\
 \Leftrightarrow \frac{2}{{x - 2}} = \frac{{ - 1}}{x}\\
 \Leftrightarrow 2x =  - x + 2\\
 \Leftrightarrow x = \frac{2}{3}\,\,\,\,\,\left( {t/m} \right)
\end{array}\)