Bài tập 2.59 trang 131 SBT Toán 12

a)

\(\begin{array}{l}
{3^{|x - 2|}} < 9 = {3^2}\\
 \Leftrightarrow |x - 2| < 2\\
 \Leftrightarrow  - 2 < x - 2 < 2\\
 \Leftrightarrow 0 < x < 4
\end{array}\)

b) 

\(\begin{array}{l}
{4^{|x + 1|}} > 16 = {4^2}\\
 \Leftrightarrow |x + 1| > 2\\
 \Leftrightarrow \left[ \begin{array}{l}
x + 1 <  - 2\\
x + 1 > 2
\end{array} \right.\\
 \Leftrightarrow \left[ \begin{array}{l}
x <  - 3\\
x > 1
\end{array} \right.
\end{array}\)

c) 

\(\begin{array}{l}
{2^{ - {x^2} + 3x}} < 4 = {2^2}\\
 \Leftrightarrow  - {x^2} + 3x < 2\\
 \Leftrightarrow  - {x^2} + 3x - 2 < 0\\
 \Leftrightarrow \left[ \begin{array}{l}
x < 1\\
x > 2
\end{array} \right.
\end{array}\)

d) 

\(\begin{array}{l}
{\left( {\frac{7}{9}} \right)^{2{x^2} - 3x}} \ge \frac{9}{7} = {\left( {\frac{7}{9}} \right)^{ - 1}}\\
 \Leftrightarrow 2{x^2} - 3x \le  - 1\\
 \Leftrightarrow 2{x^2} - 3x + 1 \le 0\\
 \Leftrightarrow x \in [12;1]
\end{array}\)

e) ĐK: \(x \ge  - 6\)

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

+) Luôn đúng với \(x \in [ - 6;0]\)
+) Nếu x > 0, ta có:
\(\begin{array}{l}
x + 6 \ge {x^2}\\
 \Leftrightarrow {x^2} - x - 6 \le 0\\
 \Leftrightarrow \left( {x - 3} \right)\left( {x + 2} \right) \le 0\\
 \Leftrightarrow x \in \left[ { - 2;3} \right]
\end{array}\)
Vậy \(x \in [ - 6;0] \cup [ - 2;3] = [ - 6;3]\)

g) 

\(\begin{array}{l}
{2^{2x - 1}} + {2^{2x - 2}} + {2^{2x - 3}} \ge 448\\
 \Leftrightarrow {2^{2x - 3}}({2^2} + 2 + 1) \ge 448\\
 \Leftrightarrow {2^{2x - 3}} \ge 64 = {2^6}\\
 \Leftrightarrow 2x - 3 \ge 6\\
 \Leftrightarrow x \ge \frac{9}{2}
\end{array}\)

h) 

\(\begin{array}{l}
{16^x} - {4^x} - 6 \le 0\\
 \Leftrightarrow {4^{2x}} - {4^x} - 6 \le 0\\
 \Leftrightarrow ({4^x} - 3)({4^x} + 2) \le 0\\
 \Leftrightarrow 4x - 3 \le 0\,\,\,({4^x} + 2 > 0,\forall x)\\
 \Leftrightarrow x \le {\log _4}3
\end{array}\)

i) 

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

-- Mod Toán 12 HỌC247