De Thi Tham Khao Ky Thi Tot Nghiep THPT Nam 2023 Mon Toan 0 ratings0% found this document useful (0 votes) 12 views23 pagesDocument Informationclick to expand document informationOriginal TitleDe Thi Tham Khao Ky Thi Tot Nghiep Thpt Nam 2023 Mon Toan Copyright© © All Rights Reserved Available FormatsPDF, TXT or read online from Scribd Share this documentShare or Embed DocumentSharing Options
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Có bao nhiêu số nguyên \(x\) thỏa mãn \({\rm{lo}}{{\rm{g}}_3}\frac{{{x^2} – 16}}{{343}} < {\rm{lo}}{{\rm{g}}_7}\frac{{{x^2} – 16}}{{27}}\) ? A. 193. B. 92. C. 186. D. 184. Lời giải: Chọn D TXĐ: \(D = \left( { – \infty ; – 4} \right) \cup \left( {4; + \infty } \right).\) Ta có: \(\begin{array}{l}{\rm{lo}}{{\rm{g}}_3}\frac{{{x^2} – 16}}{{343}} < {\rm{lo}}{{\rm{g}}_7}\frac{{{x^2} – 16}}{{27}}\\ \Leftrightarrow {\rm{lo}}{{\rm{g}}_3}7.\left[ {{\rm{lo}}{{\rm{g}}_7}\left( {{x^2} – 16} \right) – 3} \right] < {\rm{lo}}{{\rm{g}}_7}\left( {{x^2} – 16} \right) – 3{\rm{lo}}{{\rm{g}}_7}3\\ \Leftrightarrow \left( {{\rm{lo}}{{\rm{g}}_3}7 – 1} \right){\rm{.lo}}{{\rm{g}}_7}\left( {{x^2} – 16} \right) < 3{\rm{lo}}{{\rm{g}}_3}7 – 3{\rm{lo}}{{\rm{g}}_7}3\\ \Leftrightarrow {\log _7}\left( {{x^2} – 16} \right) < \frac{{3\left( {{{\log }_3}7 – {{\log }_7}3} \right)}}{{{{\log }_3}7 – 1}}\\ \Leftrightarrow {\log _7}\left( {{x^2} – 16} \right) < 3\left( {1 + {{\log }_7}3} \right)\\ \Leftrightarrow {\log _7}\left( {{x^2} – 16} \right) < {\log _7}{21^3}\\ \Leftrightarrow {x^2} – 16 < {21^3}\\ \Leftrightarrow – \sqrt {9277} < x < \sqrt {9277} \end{array}\) Kết hợp điều kiện ta có \(x \in \left\{ { – 96; – 95;…; – 5;5;…;95;96} \right\}\) . Vậy có 184 số nguyên x thỏa mãn. Sử dụng công thức: \({\log _a}\dfrac{x}{y} = {\log _a}x - {\log _a}y,\,\,{\log _a}b = \dfrac{{{{\log }_c}b}}{{{{\log }_c}a}}\) đưa về phương trình bậc nhất với ẩn \({\log _3}\left( {{x^2} - 16} \right).\) |