Bài tập 2.70 trang 133 SBT Toán 12

a)

\(\begin{array}{l}
{\left( {8,4} \right)^{\frac{{x - 3}}{{{x^2} + 1}}}} < 1\\
 \Leftrightarrow \frac{{x - 3}}{{{x^2} + 1}} < {\log _{8,4}}1 = 0\\
 \Leftrightarrow x - 3 < 0\\
 \Leftrightarrow x < 3
\end{array}\)

b)

\(\begin{array}{l}
{2^{|x - 2|}} > {4^{|x + 1|}}\\
 \Leftrightarrow {2^{|x - 2|}} > {2^{2|x + 1|}}\\
 \Leftrightarrow |x - 2| > 2|x + 1|\\
 \Leftrightarrow {x^2} - 4x + 4 > 4{x^2} + 8x + 4\\
 \Leftrightarrow 3{x^2} + 12x < 0\\
 \Leftrightarrow  - 4 < x < 0
\end{array}\)

c)

\(\begin{array}{l}
\frac{{{4^x} - {2^{x + 1}} + 8}}{{{2^{1 - x}}}} < {8^x}\\
 \Leftrightarrow {4^x} - {2^{x + 1}} + 8 < {2^{1 - x}}{.2^{3x}}\\
 \Leftrightarrow {2^{2x}} - {2.2^x} + 8 < {2^{2x + 1}}\\
 \Leftrightarrow  - {2^{2x}} - {2.2^x} + 8 < 0\\
 \Leftrightarrow \left[ \begin{array}{l}
{2^x} <  - 4\,\,\left( l \right)\\
{2^x} > 2
\end{array} \right.\\
 \Leftrightarrow x > 1
\end{array}\)

d)

\(\begin{array}{l}
\frac{1}{{{3^x} + 5}} \le \frac{1}{{{3^{x + 1}} - 1}}\\
 \Leftrightarrow \frac{1}{{{3^x} + 5}} - \frac{1}{{{3^{x + 1}} - 1}} \le 0\\
 \Leftrightarrow \frac{{{{3.3}^x} - 1 - ({3^x} + 5)}}{{({3^x} + 5)({{3.3}^x} - 1)}} \le 0\\
 \Leftrightarrow \frac{{{{2.3}^x} - 6}}{{({3^x} + 5)({{3.3}^x} - 1)}} \le 0\\
 \Leftrightarrow \frac{1}{3} < {3^x} \le 3 \Leftrightarrow  - 1 < x \le 1
\end{array}\)

-- Mod Toán 12 HỌC247