\[{5.4^x} + {2.25^x} = {7.10^x}\]

Chia cae 2 vế cho \[{4^x}\] ta có:

\[\begin{array}{l}
5 + 2.\frac{{{{25}^x}}}{{{4^x}}} = 7.\frac{{{{10}^x}}}{{{4^x}}}\\
\Leftrightarrow 5 + 2.{\left( {\frac{5}{2}} \right)^{2x}} = 7.{\left( {\frac{5}{2}} \right)^x}\\
dat:{\left( {\frac{5}{2}} \right)^x} = t\left( {t \ge 0} \right)\\
= > 5 + 2{t^2} = 7t\\
\Leftrightarrow 2{t^2} - 2t - 5t + 5 = 0\\
\Leftrightarrow 2t\left( {t - 1} \right) - 5\left( {t - 1} \right) = 0\\
\Leftrightarrow \left( {t - 1} \right)\left( {2t - 5} \right) = 0\\
\Leftrightarrow \left[ \begin{array}{l}
t - 1 = 0\\
2t - 5 = 0
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
t = 1\\
t = \frac{5}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
{\left( {\frac{5}{2}} \right)^x} = 1\\
{\left( {\frac{5}{2}} \right)^x} = \frac{5}{2}
\end{array} \right.\\
\Leftrightarrow \left[ \begin{array}{l}
x = 0\\
x = 1
\end{array} \right.\\
vay:S = \{ 0;1\}
\end{array}\]