Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne - 1;x \ne - 3\\
\dfrac{{4x}}{{{x^2} + 4x + 3}} - 1 = 6.\left( {\dfrac{1}{{x + 3}} - \dfrac{1}{{2x + 2}}} \right)\\
\Rightarrow \dfrac{{4x - {x^2} - 4x - 3}}{{\left( {x + 1} \right)\left( {x + 3} \right)}} = \dfrac{6}{{x + 3}} - \dfrac{3}{{x + 1}}\\
\Rightarrow \dfrac{{ - {x^2} - 3}}{{\left( {x + 1} \right)\left( {x + 3} \right)}} = \dfrac{{6\left( {x + 1} \right) - 3\left( {x + 3} \right)}}{{\left( {x + 1} \right)\left( {x + 3} \right)}}\\
\Rightarrow - {x^2} - 3 = 6x + 6 - 3x - 3\\
\Rightarrow {x^2} + 3x + 6 = 0\left( {vn} \right)\\
\text{Vậy pt vô nghiệm}\\
b)Dkxd:x \ne 1\\
\dfrac{{ - 1}}{{1 - x}} + \dfrac{{2{x^2} - 5}}{{{x^3} - 1}} = \dfrac{4}{{{x^2} + x + 1}}\\
\Rightarrow \dfrac{1}{{x - 1}} + \dfrac{{2{x^2} - 5}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{4}{{{x^2} + x + 1}}\\
\Rightarrow \dfrac{{{x^2} + x + 1 + 2{x^2} - 5}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \dfrac{{4\left( {x - 1} \right)}}{{{x^2} + x + 1}}\\
\Rightarrow 3{x^2} + x - 4 = 4x - 4\\
\Rightarrow 3{x^2} - 3x = 0\\
\Rightarrow 3x\left( {x - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\left( {ktm} \right)\\
\text{Vậy}\,x = 0\\
c)Dkxd:x \ne \dfrac{1}{3};x \ne - \dfrac{1}{3}\\
\dfrac{{12x + 1}}{{6x - 2}} - \dfrac{{9x - 5}}{{3x - 1}} = \dfrac{{108x - 36{x^2} - 9}}{{4\left( {9{x^2} - 1} \right)}}\\
\Rightarrow \dfrac{{12x + 1 - 18x + 10}}{{2\left( {3x - 1} \right)}} = \dfrac{{108x - 36{x^2} - 9}}{{4\left( {3x - 1} \right)\left( {3x + 1} \right)}}\\
\Rightarrow \dfrac{{\left( {11 - 6x} \right).2.\left( {3x + 1} \right)}}{{4\left( {3x - 1} \right)\left( {3x + 1} \right)}} = \dfrac{{108x - 36{x^2} - 9}}{{\left( {3x - 1} \right)\left( {3x + 1} \right)}}\\
\Rightarrow - 36{x^2} + 66x - 12x + 22 = 108x - 36{x^2} - 9\\
\Rightarrow 108x - 66x + 12x = 22 + 9\\
\Rightarrow 54x = 31\\
\Rightarrow x = \dfrac{{31}}{{54}}\left( {tmdk} \right)\\
\text{Vậy}\,x = \dfrac{{31}}{{54}}
\end{array}$
