Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne 1;x \ne - 1\\
\dfrac{{2x + 1}}{{x - 1}} = \dfrac{{5\left( {x - 1} \right)}}{{x + 1}}\\
\Rightarrow \left( {2x + 1} \right)\left( {x + 1} \right) = 5{\left( {x - 1} \right)^2}\\
\Rightarrow 2{x^2} + 3x + 1 = 5{x^2} - 10x + 5\\
\Rightarrow 3{x^2} - 13x + 4 = 0\\
\Rightarrow 3{x^2} - 12x - x + 4 = 0\\
\Rightarrow \left( {x - 4} \right)\left( {3x - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 4\left( {tm} \right)\\
x = \dfrac{1}{3}\left( {tm} \right)
\end{array} \right.\\
b)\dfrac{{x - 3}}{{x - 2}} + \dfrac{{x - 2}}{{x - 4}} = - 1\left( {x \ne 2;x \ne 4} \right)\\
\Rightarrow \dfrac{{\left( {x - 3} \right)\left( {x - 4} \right) + {{\left( {x - 2} \right)}^2}}}{{\left( {x - 2} \right)\left( {x - 4} \right)}} = - 1\\
\Rightarrow {x^2} - 7x + 12 + {x^2} - 4x + 4 = - {x^2} + 6x - 8\\
\Rightarrow 2{x^2} - 11x + 16 = - {x^2} + 6x - 8\\
\Rightarrow 3{x^2} - 17x + 24 = 0\\
\Rightarrow 3{x^2} - 9x - 8x + 24 = 0\\
\Rightarrow \left( {x - 3} \right)\left( {3x - 8} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 3\left( {tm} \right)\\
x = \dfrac{8}{3}\left( {tm} \right)
\end{array} \right.\\
c)Dkxd:x \ne 1\\
\dfrac{1}{{x - 1}} + \dfrac{{2{x^2} - 5}}{{{x^3} - 1}} = \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)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
\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\left( {tm} \right)\\
x = 1\left( {ktm} \right)
\end{array} \right.\\
Vay\,x = 0\\
d)Dk:x \ne 3;x \ne \dfrac{{ - 7}}{2};x \ne - 3\\
\dfrac{{13}}{{\left( {x - 3} \right)\left( {2x + 7} \right)}} + \dfrac{1}{{2x + 7}} = \dfrac{6}{{{x^2} - 9}}\\
\Rightarrow \dfrac{{13\left( {x + 3} \right) + {x^2} - 9}}{{\left( {x - 3} \right)\left( {x + 3} \right)\left( {2x + 7} \right)}} = \dfrac{{6\left( {2x + 7} \right)}}{{\left( {x - 3} \right)\left( {x + 3} \right)\left( {2x + 7} \right)}}\\
\Rightarrow 13x + 39 + {x^2} - 9 = 12x + 42\\
\Rightarrow {x^2} + x - 12 = 0\\
\Rightarrow \left( {x - 3} \right)\left( {x + 4} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 3\left( {ktm} \right)\\
x = - 4\left( {tm} \right)
\end{array} \right.\\
Vậy\,x = - 4
\end{array}$
