a) Đặt \(\displaystyle t = {3^x} > 0\) ta được: \(\displaystyle {t^2} - t - 6 = 0 \Leftrightarrow \left[ \begin{array}{l}t = 3\left( {TM} \right)\\t = - 2\left( {KTM} \right)\end{array} \right.\)
Suy ra \(\displaystyle {3^x} = 3 \Leftrightarrow x = 1\).
b) Đặt \(\displaystyle t = {e^x}(t > 0)\), ta có phương trình \(\displaystyle {t^2} - 3t - 4 + \frac{{12}}{t} = 0\)
\(\displaystyle \Rightarrow {t^3} - 3{t^2} - 4t + 12 = 0\)\(\displaystyle \Leftrightarrow (t - 2)(t + 2)(t - 3) = 0\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}t = 2\\t = - 2(l)\\t = 3\end{array} \right.\)
Do đó \(\displaystyle \left[ \begin{array}{l}{e^x} = 2\\{e^x} = 3\end{array} \right.\) hay \(\displaystyle \left[ \begin{array}{l}x = \ln 2\\x = \ln 3\end{array} \right.\)
c) \(\displaystyle {3.4^x} + \frac{1}{3}{.9^{x + 2}} = {6.4^{x + 1}} - \frac{1}{2}{.9^{x + 1}}\)\(\displaystyle \Leftrightarrow {3.4^x} + \frac{1}{3}{.9^x}{.9^2} = {6.4^x}.4 - \frac{1}{2}{.9^x}.9\)
\(\displaystyle \Leftrightarrow {3.4^x} + {27.9^x} = {24.4^x} - \frac{9}{2}{.9^x}\)\(\displaystyle \Leftrightarrow \frac{{63}}{2}{.9^x} = {21.4^x}\) \(\displaystyle \Leftrightarrow {63.9^x} = {42.4^x}\) \(\displaystyle \Leftrightarrow {\left( {\frac{9}{4}} \right)^x} = \frac{2}{3}\)
\(\displaystyle \Leftrightarrow {\left( {\frac{3}{2}} \right)^{2x}} = {\left( {\frac{3}{2}} \right)^{ - 1}}\)\(\displaystyle \Leftrightarrow 2x = - 1 \Leftrightarrow x = - \frac{1}{2}\)
d) \(\displaystyle {2^{{x^2} - 1}} - {3^{{x^2}}} = {3^{{x^2} - 1}} - {2^{{x^2} + 2}}\) \(\displaystyle \Leftrightarrow \frac{1}{2}{.2^{{x^2}}} - {3^{{x^2}}} = \frac{1}{3}{.3^{{x^2}}} - {4.2^{{x^2}}}\)\(\displaystyle \Leftrightarrow \frac{9}{2}{.2^{{x^2}}} = \frac{4}{3}{.3^{{x^2}}} \Leftrightarrow {\left( {\frac{2}{3}} \right)^{{x^2}}} = {\left( {\frac{2}{3}} \right)^3}\)
\(\displaystyle \Leftrightarrow {x^2} = 3 \Leftrightarrow \left[ \begin{array}{l}x = \sqrt 3 \\x = - \sqrt 3 \end{array} \right.\)
