a) \(\displaystyle {16.2^x} + {4.2^x} = {5.5^x} + {3.5^x}\)\(\displaystyle \Leftrightarrow {20.2^x} = {8.5^x}\) \(\displaystyle \Leftrightarrow {\left( {\frac{2}{5}} \right)^x} = {\left( {\frac{2}{5}} \right)^1} \Leftrightarrow x = 1\)
b) \(\displaystyle {16.7^x} - {16.5^{2x}} = 0 \Leftrightarrow {7^x} = {5^{2x}}\)\(\displaystyle \Leftrightarrow {\left( {\frac{7}{{25}}} \right)^x} = {\left( {\frac{7}{{25}}} \right)^0} \Leftrightarrow x = 0\)
c) Chia hai vế cho \(\displaystyle {12^x}({12^x} > 0)\), ta được: \(\displaystyle 4{\left( {\frac{3}{4}} \right)^x} + 1 - 3{\left( {\frac{4}{3}} \right)^x} = 0\)
Đặt \(\displaystyle t = {\left( {\frac{3}{4}} \right)^x} > 0\), ta có phương trình: \(\displaystyle 4t + 1 - \frac{3}{t} = 0\)\(\displaystyle \Leftrightarrow 4{t^2} + t - 3 = 0\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}t = - 1\left( {KTM} \right)\\t = \frac{3}{4}\left( {TM} \right)\end{array} \right.\)
Do đó \(\displaystyle {\left( {\frac{3}{4}} \right)^x} = {\left( {\frac{3}{4}} \right)^1} \Leftrightarrow x = 1\) .
Vậy \(\displaystyle x = 1\).
d) Đặt \(\displaystyle t = {2^x}(t > 0)\) , ta có phương trình:
\(\displaystyle - {t^3} + 2{t^2} + t - 2 = 0\)\(\displaystyle \Leftrightarrow (t - 1)(t + 1)(2 - t) = 0\) \(\displaystyle \Leftrightarrow \left[ \begin{array}{l}t = 1\left( {TM} \right)\\t = - 1\left( {KTM} \right)\\t = 2\left( {TM} \right)\end{array} \right.\)
Do đó \(\displaystyle \left[ \begin{array}{l}{2^x} = 1\\{2^x} = 2\end{array} \right. \Leftrightarrow \left[ \begin{array}{l}x = 0\\x = 1\end{array} \right.\)
Vậy phương trình có nghiệm \(\displaystyle x = 1\), \(\displaystyle x = 0\).
