Đáp án:

\[S = \left( { - \infty ;{{\log }_2}\frac{{\sqrt {21}  - 3}}{2}} \right) \cup \left( {{{\log }_3}\frac{{1 + \sqrt {13} }}{2};1} \right)\]

Giải thích các bước giải:

 Ta có:

\(\begin{array}{l}
{27^x} - {27^{1 - x}} - 16.\left( {{3^x} - \frac{3}{{{3^x}}}} \right) + 6 < 0\\
 \Leftrightarrow \left( {{{27}^x} - \frac{{27}}{{{{27}^x}}}} \right) - 16\left( {{3^x} - \frac{3}{{{3^x}}}} \right) + 6 < 0\,\,\,\,\,\left( 1 \right)\\
t = {3^x} - \frac{3}{{{3^x}}} \Rightarrow {t^3} = {\left( {{3^x} - \frac{3}{{{3^x}}}} \right)^3} = {27^x} - {3.9^x}.\frac{3}{{{3^x}}} + {3.3^x}.\frac{9}{{{9^x}}} - \frac{{27}}{{{{27}^x}}}\\
 \Leftrightarrow {t^3} = \left( {{{27}^x} - \frac{{27}}{{{{27}^x}}}} \right) - {3.3^x}.\frac{3}{{{3^x}}}\left( {{3^x} - \frac{3}{{{3^x}}}} \right)\\
 \Leftrightarrow {t^3} = \left( {{{27}^x} - \frac{{27}}{{{{27}^x}}}} \right) - 3.3.t\\
 \Leftrightarrow {27^x} - \frac{{27}}{{{{27}^x}}} = {t^3} + 9t\\
\left( 1 \right) \Leftrightarrow {t^3} + 9t - 16t + 6 < 0\\
 \Leftrightarrow {t^3} - 7t + 6 < 0\\
 \Leftrightarrow \left[ \begin{array}{l}
t <  - 3\\
1 < t < 2
\end{array} \right.\\
TH1:\,\,\,t <  - 3\\
 \Leftrightarrow {3^x} - \frac{3}{{{3^x}}} + 3 < 0\\
 \Leftrightarrow {\left( {{3^x}} \right)^2} + 3.\left( {{3^x}} \right) - 3 < 0\\
 \Leftrightarrow \frac{{ - 3 - \sqrt {21} }}{2} < {3^x} < \frac{{ - 3 + \sqrt {21} }}{2}\\
 \Leftrightarrow x < {\log _3}\frac{{\sqrt {21}  - 3}}{2}\\
TH2:\,\,\,1 < t < 2\\
t > 1 \Leftrightarrow {3^x} - \frac{3}{{{3^x}}} - 1 > 0\\
 \Leftrightarrow {\left( {{3^x}} \right)^2} - \left( {{3^x}} \right) - 3 > 0\\
 \Leftrightarrow {3^x} > \frac{{1 + \sqrt {13} }}{2}\\
 \Leftrightarrow x > {\log _3}\frac{{1 + \sqrt {13} }}{2}\\
t < 2 \Leftrightarrow {3^x} - \frac{3}{{{3^x}}} < 2\\
 \Leftrightarrow {\left( {{3^x}} \right)^2} - {2.3^x} - 3 < 0\\
 \Leftrightarrow  - 1 < {3^x} < 3\\
 \Leftrightarrow x < 1\\
 \Rightarrow {\log _3}\frac{{1 + \sqrt {13} }}{2} < x < 1\\
 \Rightarrow S = \left( { - \infty ;{{\log }_2}\frac{{\sqrt {21}  - 3}}{2}} \right) \cup \left( {{{\log }_3}\frac{{1 + \sqrt {13} }}{2};1} \right)
\end{array}\)