$\begin{array}{l}
(m + 1){4^x} + (3m - 2){2^{x + 1}} - 3m + 1 = 0\,\,\,\,(*)\\
\Leftrightarrow (m + 1){2^{2x}} + 2(3m - 2){2^x} - 3m + 1 = 0\,\\
Dat\,\,\,{2^x} = t > 0\,\, \Rightarrow (m + 1){t^2} + 2(3m - 2)t - 3m + 1 = 0\,\,\,\,(1)
\end{array}$
\((*)\) có hai nghiệm trái dấu \({x_1},{x_2}\, \Leftrightarrow (1)\,\) có hai nghiệm \({t_1},{t_2}\) sao cho \({t_1} < 1 < {t_2}\)
$\begin{array}{l}
\Leftrightarrow \left\{ \begin{array}{l}
\Delta ' > 0\\
\left( {{t_1} - 1} \right)\left( {{t_2} - 1} \right) < 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
{(3m - 2)^2} - (m + 1)( - 3m + 1) > 0\\
{t_1}{t_2} - ({t_1} + {t_2}) + 1 < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
9{m^2} - 12m + 4 + 3{m^2} + 3m - m - 1 > 0\\
\frac{{ - 3m + 1}}{{m + 1}} + \frac{{2(3m - 2)}}{{m + 1}} + 1 < 0
\end{array} \right.\\
\Leftrightarrow \left\{ \begin{array}{l}
12{m^2} - 10m + 3 > 0\,\,\,(Dung)\\
\frac{{ - 3m + 1 + 6m - 4 + m + 1}}{{m + 1}} < 0
\end{array} \right.\\
\Leftrightarrow \frac{{4m - 2}}{{m + 1}} < 0 \Leftrightarrow - 1 < m < \frac{1}{2}
\end{array}$
