\({\log _2}\left( {{{\log }_{\frac{1}{3}}}\frac{{3x - 7}}{{x + 3}}} \right) \ge 0 \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{\frac{{3x - 7}}{{x + 3}} > 0}\\{ log{ _{\frac{1}{3}}}\frac{{3x - 7}}{{x + 3}} > 0}\\{ log{ _{\frac{1}{3}}}\frac{{3x - 7}}{{x + 3}} \ge 1}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{\frac{{3x - 7}}{{x + 3}} > 0}\\{\frac{{3x - 7}}{{x + 3}} < 1}\\{\frac{{3x - 7}}{{x + 3}} \le \frac{1}{3}}\end{array} \Leftrightarrow \left\{ {\begin{array}{*{20}{c}}{\frac{{3x - 7}}{{x + 3}} > 0}\\{\frac{{3x - 7}}{{x + 3}} \le \frac{1}{3}}\end{array}} \right.} \right.} \right.\)

\( \Leftrightarrow \left\{ {\begin{array}{*{20}{l}}{\frac{{3x - 7}}{{x + 3}} > 0}\\{\frac{{8(x - 3)}}{{3(x + 3)}} \le 0}\end{array}} \right.\)

\( \Rightarrow \left\{ {\begin{array}{*{20}{c}}{x \in ( - \infty ; - 3) \cup \left( {\frac{7}{3}; + \infty } \right)}\\{x \in ( - 3;3]}\end{array} \Leftrightarrow x \in \left( {\frac{7}{3};3} \right].} \right.\)

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

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

Do đó ta điền đáp án như sau

Bất phương trình \({\rm{lo}}{{\rm{g}}_2}\left( {{\rm{lo}}{{\rm{g}}_{\frac{1}{3}}}\frac{{3x - 7}}{{x + 3}}} \right) \ge 0\) có tập nghiệm là \(\left( {a;b} \right]\) (với \(a;b \in \mathbb{R}\) ). Giá trị \(P = 3a - b\) bằng (1) 4.