Bài 2.59 trang 131 SBT giải tích 12

Đề bài

Giải các bất phương trình mũ sau:

a) \(\displaystyle {3^{|x - 2|}} < 9\)            b) \(\displaystyle {4^{|x + 1|}} > 16\)

c) \(\displaystyle {2^{ - {x^2} + 3x}} < 4\)   d) \(\displaystyle {\left( {\frac{7}{9}} \right)^{2{x^2} - 3x}} \ge \frac{9}{7}\)

e) \(\displaystyle {11^{\sqrt {x + 6} }} \ge {11^x}\)

g) \(\displaystyle {2^{2x - 1}} + {2^{2x - 2}} + {2^{2x - 3}} \ge 448\)

h) \(\displaystyle {16^x} - {4^x} - 6 \le 0\)    i) \(\displaystyle \frac{{{3^x}}}{{{3^x} - 2}} < 3\)

Lời giải chi tiết

a) \(\displaystyle {3^{|x - 2|}} < {3^2} \Leftrightarrow |x - 2| < 2\)\(\displaystyle  \Leftrightarrow  - 2 < x - 2 < 2\) \(\displaystyle  \Leftrightarrow 0 < x < 4\)

b) \(\displaystyle {4^{|x + 1|}} > {4^2} \Leftrightarrow |x + 1| > 2\)\(\displaystyle  \Leftrightarrow \left[ \begin{array}{l}x + 1 > 2\\x + 1 <  - 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x > 1\\x <  - 3\end{array} \right.\)

c) \(\displaystyle {2^{ - {x^2} + 3x}} < {2^2}\)\(\displaystyle  \Leftrightarrow  - {x^2} + 3x < 2\) \(\displaystyle  \Leftrightarrow {x^2} - 3x + 2 > 0\) \(\displaystyle  \Leftrightarrow \left[ \begin{array}{l}x < 1\\x > 2\end{array} \right.\)

d) \(\displaystyle {\left( {\frac{7}{9}} \right)^{2{x^2} - 3x}} \ge {\left( {\frac{7}{9}} \right)^{ - 1}}\)\(\displaystyle  \Leftrightarrow 2{x^2} - 3x \le  - 1\) \(\displaystyle  \Leftrightarrow 2{x^2} - 3x + 1 \le 0\) \(\displaystyle  \Leftrightarrow \frac{1}{2} \le x \le 1\)

e) \(\displaystyle \sqrt {x + 6}  \ge x\)\(\displaystyle  \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x + 6 \ge 0\\x < 0\end{array} \right.\\\left\{ \begin{array}{l}x \ge 0\\x + 6 \ge {x^2}\end{array} \right.\end{array} \right.\) \(\displaystyle  \Leftrightarrow \left[ \begin{array}{l}\left\{ \begin{array}{l}x \ge  - 6\\x < 0\end{array} \right.\\\left\{ \begin{array}{l}x \ge 0\\{x^2} - x - 6 \le 0\end{array} \right.\end{array} \right.\) \(\displaystyle  \Leftrightarrow \left[ \begin{array}{l} - 6 \le x < 0\\\left\{ \begin{array}{l} - 2 \le x \le 3\\x \ge 0\end{array} \right.\end{array} \right.\)

\(\displaystyle  \Leftrightarrow \left[ \begin{array}{l} - 6 \le x < 0\\0 \le x \le 3\end{array} \right.\)\(\displaystyle  \Leftrightarrow  - 6 \le x \le 3\)

g) \(\displaystyle \frac{1}{2}{.2^{2x}} + \frac{1}{4}{.2^{2x}} + \frac{1}{8}{.2^{2x}} \ge 448\)\(\displaystyle  \Leftrightarrow {2^{2x}} \ge 512\) \(\displaystyle  \Leftrightarrow {2^{2x}} \ge {2^9} \Leftrightarrow x \ge \frac{9}{2}\)

h) Đặt \(\displaystyle t = {4^x} > 0\), ta có hệ bất phương trình:

\(\displaystyle \left\{ \begin{array}{l}{t^2} - t - 6 \le 0\\t > 0\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} - 2 \le t \le 3\\t > 0\end{array} \right.\)\(\displaystyle  \Leftrightarrow 0 < t \le 3 \Leftrightarrow 0 < {4^x} \le 3\) \(\displaystyle  \Leftrightarrow x \le {\log _4}3\)

i) \(\displaystyle \frac{{{3^x}}}{{{3^x} - 2}} - 3 < 0\)\(\displaystyle  \Leftrightarrow \frac{{ - {{2.3}^x} + 6}}{{{3^x} - 2}} < 0\) \(\displaystyle  \Leftrightarrow \frac{{{3^x} - 3}}{{{3^x} - 2}} > 0 \Leftrightarrow \left[ \begin{array}{l}{3^x} > 3\\{3^x} < 2\end{array} \right.\) \(\displaystyle  \Leftrightarrow \left[ \begin{array}{l}x > 1\\x < {\log _3}2\end{array} \right.\)