Bất phương trình log2 (log1/3 3x-7/x+3)>0

Đáp án: “4”

Giải thích

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

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

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

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

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