Đáp án:
$\begin{array}{l}
Dkxd:x \ne 1;x \ne - 3\\
\frac{{3x - 1}}{{x - 1}} - \frac{{2x + 5}}{{x + 3}} + \frac{4}{{\left( {x - 1} \right)\left( {x + 3} \right)}} = 1\\
\Rightarrow \frac{{\left( {3x - 1} \right)\left( {x + 3} \right) - \left( {2x + 5} \right)\left( {x - 1} \right) + 4}}{{\left( {x - 1} \right)\left( {x + 3} \right)}} = 1\\
\Rightarrow 3{x^2} + 9x - x - 3 - 2{x^2} + 2x - 5x + 5 + 4\\
= {x^2} + 2x - 3\\
\Rightarrow {x^2} + 5x + 6 = {x^2} + 2x - 3\\
\Rightarrow 3x = - 9\\
\Rightarrow x = - 3\left( {ktm} \right)
\end{array}$
Vậy pt vô nghiệm.
