Có bao nhiêu số nguyên $x$ thỏa mãn $\left( {{\log }_{3}}\left(...

Ta có:
Điều kiện: .
Xét $\left[ \begin{aligned}
& {{\log }_{3}}\left( {{x}^{2}}+10 \right)-{{\log }_{3}}\left( x+40 \right)=0 \\
& 32-{{2}^{x-1}}=0 \\
\end{aligned} \right.$$\Leftrightarrow \left[ \begin{aligned}
& {{\log }_{3}}\left( {{x}^{2}}+10 \right)={{\log }_{3}}\left( x+40 \right) \\
& 32={{2}^{x-1}} \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& {{x}^{2}}+10=x+40 \\
& 5=x-1 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=-5 \\
& x=6 \\
\end{aligned} \right.\left\{ \begin{aligned}
& {{\log }_{3}}\left( {{x}^{2}}+10 \right)-{{\log }_{3}}\left( x+40 \right)\ne 0 \\
& 32-{{2}^{x-1}}\ne 0 \\
\end{aligned} \right.$$\Leftrightarrow \left\{ \begin{aligned}
& x\ne -5 \\
& x\ne 6 \\
\end{aligned} \right.$.
Khi đó
$\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{3}}\left( {{x}^{2}}+10 \right)-{{\log }_{3}}\left( x+40 \right)>0 \\
& 32-{{2}^{x-1}}>0 \\
\end{aligned} \right.$$\cup \left\{ \begin{aligned}
& {{\log }_{3}}\left( {{x}^{2}}+10 \right)-{{\log }_{3}}\left( x+40 \right)<0 \\
& 32-{{2}^{x-1}}<0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& {{\log }_{3}}\left( {{x}^{2}}+10 \right)>{{\log }_{3}}\left( x+40 \right) \\
& 32>{{2}^{x-1}} \\
\end{aligned} \right.$$\cup \left\{ \begin{aligned}
& {{\log }_{3}}\left( {{x}^{2}}+10 \right)<{{\log }_{3}}\left( x+40 \right) \\
& 32<{{2}^{x-1}} \\
\end{aligned} \right.$.
$\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}+10>x+40 \\
& 5>x-1 \\
\end{aligned} \right.$$\cup \left\{ \begin{aligned}
& {{x}^{2}}+10<x+40 \\
& 5<x-1 \\
\end{aligned} \right.\cup \left\{ \begin{aligned}
& {{x}^{2}}-x-30<0 \\
& x-6>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& x<-5\cup x>6 \\
& x<6 \\
\end{aligned} \right.\Leftrightarrow x<-5-40<x\le -5\cup x=6xx\in \left\{ -39;-38;...-5;6 \right\}x+40>0\Leftrightarrow x>-40 \left( * \right)g\left( x \right)=\left( {{\log }_{3}}\left( {{x}^{2}}+10 \right)-{{\log }_{3}}\left( x+40 \right) \right)\left( 32-{{2}^{x-1}} \right)=0\Leftrightarrow \left[ \begin{aligned}
& {{\log }_{3}}\left( {{x}^{2}}+10 \right)-{{\log }_{3}}\left( x+40 \right)=0 \\
& 32-{{2}^{x-1}}=0 \\
\end{aligned} \right.\Leftrightarrow \left[ \begin{aligned}
& x=-5 \\
& x=6 \\
\end{aligned} \right.​