Giải và biện luận bất phương trình sau theo tham số $a$:       $x^{\log_a (ax)}\geq  (ax)^4$
 Với $a > 1$  ta có
\({x^{{{\log }_a}\left( {{\rm{ax}}} \right)}} \ge {\left( {{\rm{ax}}} \right)^4} \Leftrightarrow {\log _a}{{x^{{{\log }_a}\left( {{\rm{ax}}} \right)}} \ge {{\log }_a}{{\left( {{\rm{ax}}} \right)}^4}}\)
                              \(\Leftrightarrow {\log _a}\left( {{\rm{ax}}} \right).{\log _a}x \ge 4{\log _a}\left( {{\rm{ax}}} \right)\)
                              \(\Leftrightarrow {\log _a}\left( {{\rm{ax}}} \right)\left( {{{\log }_a}x - 4} \right) \ge 0\)
Đáp số:\(\left[ \begin{array}{l}
x \ge {a^4}\\
0 < x \le \frac{1}{a}
\end{array} \right.\)

Với $0 < a < 1$ ta có:
\({x^{{{\log }_a}\left( {{\rm{ax}}} \right)}} \ge {\left( {{\rm{ax}}} \right)^4} \Leftrightarrow \left( {1 + {{\log }_a}x} \right)\left( {{{\log }_a} - 4} \right) \le 0\)
                              \(\Leftrightarrow - 1 \le {\log _a}x \le 4 \Leftrightarrow {a^4} \le x \le \frac{1}{a}\)
Đáp số: ${a^4} \le x \le \frac{1}{a}$

Bạn cần đăng nhập để có thể gửi đáp án

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