Đáp án:
$\begin{array}{l}
a)Dkxd:x \ne - 1\\
\frac{{1 - x}}{{x + 1}} + 3 = \frac{{2x + 3}}{{x + 1}}\\
\Leftrightarrow \frac{{1 - x + 3\left( {x + 1} \right)}}{{x + 1}} = \frac{{2x + 3}}{{x + 1}}\\
\Rightarrow 1 - x + 3x + 3 = 2x + 3\\
\Rightarrow 2x + 4 = 2x + 3\\
\Rightarrow 4 = 3\left( {ktm} \right)\\
Vậy\,pt\,vô\,nghiệm\\
b)Dkxd:x \ne 1\\
\frac{1}{{x - 1}} + \frac{{2{x^2} - 5}}{{{x^3} - 1}} = \frac{4}{{{x^2} + x + 1}}\\
\Leftrightarrow \frac{1}{{x - 1}} + \frac{{2{x^2} - 5}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} - \frac{4}{{{x^2} + x + 1}} = 0\\
\Leftrightarrow \frac{{{x^2} + x + 1 + 2{x^2} - 5 - 4\left( {x - 1} \right)}}{{\left( {x - 1} \right)\left( {{x^2} + x + 1} \right)}} = 0\\
\Rightarrow 3{x^2} - 3x = 0\\
\Rightarrow 3x\left( {x - 1} \right) = 0\\
\Rightarrow \left[ \begin{array}{l}
x = 0\left( {tm} \right)\\
x = 1\left( {ktm} \right)
\end{array} \right.\\
Vậy\,x = 0
\end{array}$
