Đáp án:
$\begin{array}{l}
a)Dkxd:x\# - \dfrac{1}{3};x\# \dfrac{1}{3};x\# - 2\\
P = \left( {\dfrac{1}{{3x + 1}} - \dfrac{{6{x^2} + 1}}{{1 - 9{x^2}}} - \dfrac{{2x}}{{3x - 1}}} \right):\dfrac{{x + 2}}{{9{x^2} - 1}}\\
= \dfrac{{3x - 1 + 6{x^2} + 1 - 2x\left( {3x + 1} \right)}}{{\left( {3x + 1} \right)\left( {3x - 1} \right)}}.\dfrac{{\left( {3x + 1} \right)\left( {3x - 1} \right)}}{{x + 2}}\\
= \dfrac{{6{x^2} + 3x - 6{x^2} - 2x}}{1}.\dfrac{1}{{x + 2}}\\
= \dfrac{x}{{x + 2}}\\
b)\left| {2x + 1} \right| = 3\\
\Leftrightarrow \left[ \begin{array}{l}
2x + 1 = 3\\
2x + 1 = - 3
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 1\left( {tm} \right)\\
x = - 2\left( {ktm} \right)
\end{array} \right.\\
Khi:x = 1\\
\Leftrightarrow P = \dfrac{x}{{x + 2}} = \dfrac{1}{{1 + 2}} = \dfrac{1}{3}\\
c)P \le 1\\
\Leftrightarrow \dfrac{x}{{x + 2}} \le 1\\
\Leftrightarrow \dfrac{{x - x - 2}}{{x + 2}} \le 0\\
\Leftrightarrow \dfrac{{ - 2}}{{x + 2}} \le 0\\
\Leftrightarrow x + 2 > 0\\
\Leftrightarrow x > - 2\\
Vậy\,x > - 2;x\# \dfrac{{ - 1}}{3};x\# \dfrac{1}{3}
\end{array}$
