Giải thích các bước giải:
Ta có:
\(\begin{array}{l}
\frac{{2x + m - 1}}{{x + 1}} > 0\,\,\,\,\,\left( 1 \right)\,\,\,\,\,\left( {x \ne - 1} \right)\\
TH1:\,\,\,\frac{{m - 1}}{2} > 1 \Leftrightarrow m > 3\\
\Rightarrow - \frac{{m - 1}}{2} < - 1\\
\left( 1 \right) \Leftrightarrow \left[ \begin{array}{l}
x > - 1\\
x < - \frac{{m - 1}}{2}
\end{array} \right.\\
TH2:\,\,\frac{{m - 1}}{2} = 1 \Leftrightarrow m = 3\\
\left( 1 \right) \Leftrightarrow \frac{{2x + 2}}{{x + 1}} > 0 \Leftrightarrow 2 > 0,\,\,\,\,\left( {t/m,\,\,\forall x \ne - 1} \right)\\
TH3:\,\,\,\frac{{m - 1}}{2} < 1 \Leftrightarrow m < 3\\
\Rightarrow - \frac{{m - 1}}{2} > - 1\\
\left( 1 \right) \Leftrightarrow \left[ \begin{array}{l}
m > - \frac{{m - 1}}{2}\\
m < - 1
\end{array} \right.
\end{array}\)
