\[\begin{array}{l}
a)\frac{1}{{x - 1}} - \frac{{3{x^2}}}{{{x^3} - 1}} = \frac{{2x}}{{{x^2} + x + 1}}\\
dk:x \ne 1\\
\Leftrightarrow \frac{{{x^2} + x + x}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \frac{{3{x^2}}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = \frac{{2x\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}}\\
\Leftrightarrow {x^2} + x + 1 - 3{x^2} = 2{x^2} - 2x\\
\Leftrightarrow 4{x^2} - 3x - 1 = 0\\
\Leftrightarrow 4{x^2} + x - 4x - 1 = 0\\
\Leftrightarrow x\left( {4x + 1} \right) - \left( {4x + 1} \right) = 0\\
\Leftrightarrow \left( {4x + 1} \right)\left( {x - 1} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
4x + 1 = 0\\
x - 1 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = \frac{{ - 1}}{4}\left( {tm} \right)\\
x = 1\left( {ktm} \right)
\end{array} \right.\\
vay:x = \frac{{ - 1}}{2}\\
b)\frac{{x - 1}}{{x + 2}} - \frac{x}{{x - 2}} = \frac{{5x - 2}}{{4 - {x^2}}}\\
dk:x \ne \pm 2\\
\Leftrightarrow \frac{{\left( {x - 1} \right)\left( {x - 2} \right) - x\left( {x + 2} \right)}}{{\left( {x - 2} \right)\left( {x + 2} \right)}} = \frac{{2 - 5x}}{{\left( {x - 2} \right)\left( {x + 2} \right)}}\\
\Leftrightarrow {x^2} - 2x - x + 2 - {x^2} - 2x = 2 - 5x\\
\Leftrightarrow - 5x + 2 = 2 - 5x
\end{array}\]

Vậy pt  nghiệm đúng với mọi \[x \ne \pm 2\]

\[\begin{array}{l}
c)\frac{{x - 5}}{{{x^2} - 5x}} - \frac{{x - 5}}{{2{x^2} + 10x}} = \frac{{x + 25}}{{2{x^2} - 50}}\\
dk:x \ne ;x \ne \pm 5\\
\Leftrightarrow \frac{{x - 5}}{{x\left( {x - 5} \right)}} - \frac{{x - 5}}{{2x\left( {x + 5} \right)}} = \frac{{x + 25}}{{2\left( {{x^2} - 25} \right)}}\\
\Leftrightarrow \frac{1}{x} - \frac{{x - 5}}{{2x\left( {x + 5} \right)}} = \frac{{x + 25}}{{2\left( {x - 5} \right)\left( {x + 5} \right)}}\\
\Leftrightarrow \frac{{2\left( {x - 5} \right)\left( {x + 5} \right)}}{{2x\left( {x - 5} \right)\left( {x + 5} \right)}} - \frac{{{{\left( {x - 5} \right)}^2}}}{{2x\left( {x - 5} \right)\left( {x + 5} \right)}} = \frac{{x\left( {x + 25} \right)}}{{2x\left( {x - 5} \right)\left( {x + 5} \right)}}\\
= > 2\left( {{x^2} - 25} \right) - \left( {{x^2} - 10x + 25} \right) = {x^2} + 25x\\
\Leftrightarrow 2{x^2} - 50 - {x^2} + 10x - 25 - {x^2} - 25x = 0\\
\Leftrightarrow - 15x - 75 = 0\\
\Leftrightarrow x = 5\left( {ktm} \right)
\end{array}\]

vậy pt vô nghiệm