Đáp án:
$\begin{array}{l}
\left( {\frac{{2{t^3} + {t^2} - t}}{{{t^3} - 1}} - \frac{{{t^2} + t}}{{{t^2} - 1}}} \right).\frac{{{t^2} - 1}}{{2{t^2} + t - 1}}.\frac{t}{{2t - 1}}\\
= \left( {\frac{{2{t^3} + {t^2} - t}}{{{t^3} - 1}}.\frac{{{t^2} - 1}}{{2{t^2} + t - 1}} - \frac{{{t^2} + t}}{{{t^2} - 1}}.\frac{{{t^2} - 1}}{{2{t^2} + t - 1}}} \right).\frac{t}{{2t - 1}}\\
= \left[ {\frac{{t\left( {2{t^2} + t - 1} \right).\left( {t - 1} \right)\left( {t + 1} \right)}}{{\left( {t - 1} \right)\left( {{t^2} + t + 1} \right).\left( {2{t^2} + t - 1} \right)}} - \frac{{{t^2} + t}}{{2{t^2} + t - 1}}} \right].\frac{t}{{2t - 1}}\\
= \left[ {\frac{{t\left( {t + 1} \right)}}{{{t^2} + t + 1}} - \frac{{{t^2} + t}}{{2{t^2} + t - 1}}} \right].\frac{t}{{2t - 1}}\\
= \left( {{t^2} + t} \right).\left( {\frac{1}{{{t^2} + t + 1}} - \frac{1}{{2{t^2} + t - 1}}} \right).\frac{t}{{2t - 1}}\\
= \frac{{{t^3} + {t^2}}}{{2t - 1}}.\frac{{{t^2} - 2}}{{\left( {{t^2} + t + 1} \right).\left( {2{t^2} + t - 1} \right)}}
\end{array}$
