Giải thích các bước giải:
a,
ĐKXĐ: \(x \ne \pm \frac{1}{3}\)
Ta có:
\(\begin{array}{l}
M = \left( {\frac{{x - 1}}{{3x - 1}} - \frac{1}{{3x + 1}} + \frac{{8x}}{{9{x^2} - 1}}} \right):\left( {1 - \frac{{3x - 2}}{{3x + 1}}} \right)\\
= \left( {\frac{{x - 1}}{{3x - 1}} - \frac{1}{{3x + 1}} + \frac{{8x}}{{\left( {3x - 1} \right)\left( {3x + 1} \right)}}} \right):\frac{{\left( {3x + 1} \right) - \left( {3x - 2} \right)}}{{3x + 1}}\\
= \frac{{\left( {x - 1} \right)\left( {3x + 1} \right) - 1.\left( {3x - 1} \right) + 8x}}{{\left( {3x - 1} \right)\left( {3x + 1} \right)}}:\frac{3}{{3x + 1}}\\
= \frac{{3{x^2} - 2x - 1 - 3x + 1 + 8x}}{{\left( {3x - 1} \right)\left( {3x + 1} \right)}}:\frac{3}{{3x + 1}}\\
= \frac{{3{x^2} + 3x}}{{\left( {3x - 1} \right)\left( {3x + 1} \right)}}.\frac{{\left( {3x + 1} \right)}}{3}\\
= \frac{{3x\left( {x + 1} \right)\left( {3x + 1} \right)}}{{\left( {3x - 1} \right)\left( {3x + 1} \right).3}}\\
= \frac{{x\left( {x + 1} \right)}}{{3x - 1}}\\
b,\\
x = \frac{1}{2} \Rightarrow M = \dfrac{{\frac{1}{2}\left( {1 + \frac{1}{2}} \right)}}{{3.\frac{1}{2} - 1}} = \frac{3}{2}
\end{array}\)