Bất phương trình log 2 ( log 1/3 (3x − 7)/( x + 3) ) ≥ 0 có tập nghiệm là ( a ; b ] . Tính giá trị P = 3a − b .

\({\log _2}\left( {{{\log }_{\frac{1}{3}}}\frac{{3x - 7}}{{x + 3}}} \right) \ge 0\)\[ \Leftrightarrow \left\{ \begin{array}{l}\frac{{3x - 7}}{{x + 3}} > 0\\{\log _{\frac{1}{3}}}\frac{{3x - 7}}{{x + 3}} > 0\\{\log _{\frac{1}{3}}}\frac{{3x - 7}}{{x + 3}} \ge 1\end{array} \right.\]\[ \Leftrightarrow \left\{ \begin{array}{l}\frac{{3x - 7}}{{x + 3}} > 0\\\frac{{3x - 7}}{{x + 3}} < 1\\\frac{{3x - 7}}{{x + 3}} \le \frac{1}{3}\end{array} \right.\]\[ \Leftrightarrow \left\{ \begin{array}{l}\frac{{3x - 7}}{{x + 3}} > 0\\\frac{{3x - 7}}{{x + 3}} \le \frac{1}{3}\end{array} \right.\]\[ \Leftrightarrow \left\{ \begin{array}{l}\frac{{3x - 7}}{{x + 3}} > 0\\\frac{{8\left( {x - 3} \right)}}{{3\left( {x + 3} \right)}} \le 0\end{array} \right.\]\[ \Leftrightarrow \left\{ \begin{array}{l}x \in \left( { - \infty ;\, - 3} \right) \cup \left( {\frac{7}{3};\, + \infty } \right)\\x \in \left( { - 3;3} \right]\end{array} \right. \Leftrightarrow x \in \left( {\frac{7}{3};\,3} \right]\].

Suy ra \(a = \frac{7}{3}\); \(b = 3\).

Vậy \(P = 3a - b = 3.\frac{7}{3} - 3 = 4\).