Đáp án:
$\begin{array}{l}
Dkxd:x\# \dfrac{1}{2};x\# - \dfrac{1}{2}\\
\dfrac{3}{{2x - 1}} + 1 = \dfrac{{2x - 1}}{{2x + 1}}\\
= > \dfrac{3}{{2x - 1}} + 1 = \dfrac{{2x + 1 - 2}}{{2x + 1}}\\
= > \dfrac{3}{{2x - 1}} + 1 = 1 - \dfrac{2}{{2x + 1}}\\
= > \dfrac{3}{{2x - 1}} = \dfrac{{ - 2}}{{2x + 1}}\\
= > 3.\left( {2x + 1} \right) = - 2.\left( {2x - 1} \right)\\
= > 6x + 3 = - 4x + 2\\
= > 10x = - 1\\
= > x = - \dfrac{1}{{10}}\left( {tmdk} \right)\\
Vậy\,x = - \dfrac{1}{{10}}
\end{array}$
