Cho phương trình : $\frac{m}{4^x}-\frac{2m+1}{2^x}+m+4=0               (1)$. Tìm các giá trị của $m$ sao cho phương trình có $2 $ nghiệm ${x_1},\,{x_2}$ thỏa mãn điều kiện  $ - 1 < {x_1} < 0 < {x_2}$
Đặt $t = \frac{1}{{{2^x}}} = {2^{ - x}},\,t > 0$
$\begin{array}{l}
\left( 1 \right) \Leftrightarrow m{t^2} - \left( {2m + 1} \right)t + m + 4 = 0 &  & \left( 2 \right)\\
\end{array}$
Ta có:
$ - 1 < {x_1} < 0 < {x_2}  \Leftrightarrow 1 >  - {x_1} >  - {x_2} $
$ \Leftrightarrow \left( 2 \right)$ có $2$ nghiệm ${t_1},\,{t_2}$ thỏa mãn $0<t_{2}<1<t_{1}<2                    (3)$   
Với ${t_2} = {2^{ - {x_2}}},\,{t_1} = {2^{ - {x_1}}}$. Đặt vế trái của $(2)$ là $f\left( t \right)$
$(3) \Leftrightarrow \left\{ \begin{array}{l}
m.f\left( 1 \right) < 0\\
m.f\left( 2 \right) > 0
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
3m < 0\\
m\left( {m + 2} \right) > 0
\end{array} \right. \\\Leftrightarrow \left\{ \begin{array}{l}
m < 0\\
\left[ \begin{array}{l}
m <  - 2\\
m > 0
\end{array} \right.
\end{array} \right. \Leftrightarrow m <  - 2$

Vậy với $m <  - 2$ thì $(1)$ có $2$  nghiệm thỏa mãn $ - 1 < {x_1} < 0 < {x_2}$

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