a) Ta có \(\left( {2 + \sqrt 3 } \right)\left( {2 - \sqrt 3 } \right) = 1\) nên \(2 - \sqrt 3 = {1 \over {2 + \sqrt 3 }} = {\left( {2 + \sqrt 3 } \right)^{ - 1}}\)
Do đó \({\left( {2 + \sqrt 3 } \right)^{2x}} = 2 - \sqrt 3 \Leftrightarrow {\left( {2 + \sqrt 3 } \right)^{2x}} = {\left( {2 + \sqrt 3 } \right)^{ - 1}} \Leftrightarrow 2x = - 1 \Leftrightarrow x = - {1 \over 2}\)
Vậy tập nghiệm phương trình là \(S = \left\{ { - {1 \over 2}} \right\}\)
b)
\({2^{{x^2} - 3x + 2}} = 4 \Leftrightarrow {2^{{x^2} - 3x + 2}} = {2^2} \Leftrightarrow {x^2} - 3x + 2 = 2 \Leftrightarrow {x^2} - 3x = 0 \Leftrightarrow \left[ \matrix{ x = 0 \hfill \cr x = 3 \hfill \cr} \right.\)
Vậy \(S = \left\{ {0;3} \right\}\)c)
\(\eqalign{
& {2.3^{x + 1}} - {6.3^{x - 1}} - {3^x} = 9 \Leftrightarrow {6.3^x} - {6 \over 3}{.3^x} - {3^x} = 9 \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow {3.3^x} = 9 \Leftrightarrow {3^x} = 3 \Leftrightarrow x = 1 \cr} \)
vậy \(S = \left\{ 1 \right\}\)d)
\(\eqalign{
& {\log _3}\left( {{3^x} + 8} \right) = 2 + x \Leftrightarrow {3^x} + 8 = {3^{2 + x}} \Leftrightarrow {3^x} + 8 = {9.3^x} \cr
& \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, \Leftrightarrow {8.3^x} = 8 \Leftrightarrow {3^x} = 1 \Leftrightarrow x = 0 \cr} \)
Vậy \(S = \left\{ 0 \right\}\)
