Câu hỏi:
Có bao nhiêu số nguyên \(x\) thỏa mãn \(\left( {{9^{{x^2}}} – {{27}^x}} \right)\left[ {{{\log }_{\frac{1}{2}}}\left( {x + 2022} \right) + 1} \right] \ge 0\)?
A. \(2020\).
B. \(2022\).
C. \(5\).
D. \(4\).
GY:
Tác giả: Hồ Hữu Tình
Điều kiện \(x > – 2022\,\left( * \right).\)
Ta có
\(\left( {{9^{{x^2}}} – {{27}^x}} \right)\left[ {{{\log }_{\frac{1}{2}}}\left( {x + 2022} \right) + 1} \right] \ge 0 \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}{9^{{x^2}}} – {27^x} \ge 0\\{\log _{\frac{1}{2}}}\left( {x + 2022} \right) + 1 \ge 0\end{array} \right.\,\,\,\,\left( 1 \right)\\\left\{ \begin{array}{l}{9^{{x^2}}} – {27^x} \le 0\\{\log _{\frac{1}{2}}}\left( {x + 2022} \right) + 1 \le 0\end{array} \right.\,\,\,\,\left( 2 \right)\end{array} \right.\)
Khi đó
\(\left( 1 \right) \Leftrightarrow \left\{ \begin{array}{l}{3^{2{x^2}}} \ge {3^{3x}}\\{\log _{\frac{1}{2}}}\left( {x + 2022} \right) \ge – 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2{x^2} \ge 3x\\x + 2022 \le {\left( {\frac{1}{2}} \right)^{ – 1}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2{x^2} – 3x \ge 0\\x + 2022 \le 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\left[ \begin{array}{l}x \ge \frac{3}{2}\\x \le 0\end{array} \right.\\x \le – 2020\end{array} \right. \Leftrightarrow x \le – 2020\)\(\left( 2 \right) \Leftrightarrow \left\{ \begin{array}{l}{3^{2{x^2}}} \le {3^{3x}}\\{\log _{\frac{1}{2}}}\left( {x + 2022} \right) \le – 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2{x^2} \le 3x\\x + 2022 \ge {\left( {\frac{1}{2}} \right)^{ – 1}}\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}2{x^2} – 3x \le 0\\x + 2022 \ge 2\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}0 \le x \le \frac{3}{2}\\x \ge – 2020\end{array} \right. \Leftrightarrow 0 \le x \le \frac{3}{2}\)
Kết hợp điều kiện \(\left( * \right)\) ta có tất cả \(4\) giá trị nguyên của \(x\) là \( – 2021;\,\, – 2020;\,\,0;\,\,1\).
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