Biết $x=\dfrac{9}{4}$ là một nghiệm của bất phương trình ${{\log...

Cách 1: Trường hợp 1: \)">a>1{{\log }_{a}}\left( {{x}^{2}}-x-2 \right)>{{\log }_{a}}\left( -{{x}^{2}}+2x+3 \right)\Leftrightarrow \left\{ \begin{aligned}
& -{{x}^{2}}+2x+3>0 \\
& {{x}^{2}}-x-2>-{{x}^{2}}+2x+3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -1<x<3 \\
& 2{{x}^{2}}-3x-5>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -1<x<3 \\
& \left[ \begin{aligned}
& x<-1 \\
& x>\dfrac{5}{2} \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \dfrac{5}{2}<x<30<a<1{{\log }_{a}}\left( {{x}^{2}}-x-2 \right)>{{\log }_{a}}\left( -{{x}^{2}}+2x+3 \right)\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}-x-2>0 \\
& {{x}^{2}}-x-2<-{{x}^{2}}+2x+3 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x<-1 \\
& x>2 \\
\end{aligned} \right. \\
& 2{{x}^{2}}-3x-5<0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& \left[ \begin{aligned}
& x<-1 \\
& x>2 \\
\end{aligned} \right. \\
& -1<x<\dfrac{5}{2} \\
\end{aligned} \right.\Leftrightarrow 2<x<\dfrac{5}{2}x=\dfrac{9}{4}\in \left( 2;\dfrac{5}{2} \right)\left( * \right)T=\left( 2;\dfrac{5}{2} \right)x=\dfrac{9}{4}{{\log }_{a}}\left( {{x}^{2}}-x-2 \right)>{{\log }_{a}}\left( -{{x}^{2}}+2x+3 \right)\Leftrightarrow {{\log }_{a}}\dfrac{13}{16}>{{\log }_{a}}\dfrac{39}{16}\Leftrightarrow 0<a<10<a<1{{\log }_{a}}\left( {{x}^{2}}-x-2 \right)>{{\log }_{a}}\left( -{{x}^{2}}+2x+3 \right)\Leftrightarrow \left\{ \begin{aligned}
& {{x}^{2}}-x-2<-{{x}^{2}}+2x+3 \\
& {{x}^{2}}-x-2>0 \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& 2{{x}^{2}}-3x-5<0 \\
& \left[ \begin{aligned}
& x<-1 \\
& x>2 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow \left\{ \begin{aligned}
& -1<x<\dfrac{5}{2} \\
& \left[ \begin{aligned}
& x<-1 \\
& x>2 \\
\end{aligned} \right. \\
\end{aligned} \right.\Leftrightarrow 2<x<\dfrac{5}{2}\left( * \right)T=\left( 2;\dfrac{5}{2} \right)$ .