Cho phương trình ${{4}^{x}}-(m+3){{2}^{x}}+8=0$ ( $m$ là tham số)...

Đặt: $t={{2}^{x}}$.
Phương trình có dạng: ${{t}^{2}}-(m+3)t+8=0$ có hai nghiệm dương phân biệt
$\Leftrightarrow \left\{ \begin{aligned}
& \Delta >0 \\
& {{t}_{1}}+{{t}_{2}}>0 \\
& {{t}_{1}}{{t}_{2}}>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \Delta ={{m}^{2}}+6m-23>0 \\
& m+3>0 \\
& 8>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& m>-3 \\
& \left[ \begin{aligned}
& m>4\sqrt{2}-3 \\
& m<-4\sqrt{2}-3 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow m>4\sqrt{2}-3.$
${{2}^{{{x}_{1}}}}={{t}_{1}}\Rightarrow {{x}_{1}}=\log _{2}^{{}}{{t}_{1}}$ ; ${{2}^{{{x}_{2}}}}={{t}_{2}}\Rightarrow {{x}_{2}}={{\log }_{2}}{{t}_{2}}.$
Suy ra ${{x}_{1}}+{{x}_{2}}={{\log }_{2}}{{t}_{1}}+{{\log }_{2}}{{t}_{2}}={{\log }_{2}}\left( {{t}_{1}}{{t}_{2}} \right)={{\log }_{2}}8=3.$
Ta có: $\left( {{x}_{1}}+3 \right)\left( {{x}_{2}}+3 \right)=8$ $\Leftrightarrow {{x}_{1}}.{{x}_{2}}+3({{x}_{1}}+{{x}_{2}})+9=8\Leftrightarrow {{x}_{1}}.{{x}_{2}}=-10.$
Suy ra $\left[ \begin{aligned}
& {{x}_{1}}=-2 \\
& {{x}_{2}}=5 \\
\end{aligned} \right.\Rightarrow \left[ \begin{aligned}
& {{t}_{1}}=\dfrac{1}{4} \\
& {{t}_{2}}=32 \\
\end{aligned} \right..$
${{t}_{1}}+{{t}_{2}}=m+3\Rightarrow m+3=32,25\Leftrightarrow m=29,25$.