Bất phương trình ${{\log }_{2}}\left( {{\log...

${{\log }_{2}}\left( {{\log }_{\dfrac{1}{3}}}\dfrac{3x-7}{x+3} \right)\ge 0$ $\Leftrightarrow \left\{ \begin{aligned}
& \dfrac{3x-7}{x+3}>0 \\
& {{\log }_{\dfrac{1}{3}}}\dfrac{3x-7}{x+3}>0 \\
& {{\log }_{\dfrac{1}{3}}}\dfrac{3x-7}{x+3}\ge 1 \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& \dfrac{3x-7}{x+3}>0 \\
& \dfrac{3x-7}{x+3}<1 \\
& \dfrac{3x-7}{x+3}\le \dfrac{1}{3} \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& \dfrac{3x-7}{x+3}>0 \\
& \dfrac{3x-7}{x+3}\le \dfrac{1}{3} \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& \dfrac{3x-7}{x+3}>0 \\
& \dfrac{8\left( x-3 \right)}{3\left( x+3 \right)}\le 0 \\
\end{aligned} \right. $ $ \Leftrightarrow \left\{ \begin{aligned}
& x\in \left( -\infty ; -3 \right)\cup \left( \dfrac{7}{3}; +\infty \right) \\
& \dfrac{8\left( x-3 \right)}{3\left( x+3 \right)}<0x\in \left[ -3; 3 \right] \\
\end{aligned} \right.\Leftrightarrow x\in \left( \dfrac{7}{3}; 3 \right]$.
Suy ra $a=\dfrac{7}{3}$ ; $b=3$. Vậy $P=3a-b=3.\dfrac{7}{3}-3=4$.