${5.4^x} + {2.25^x} = {7.10^x}$

Chia cae 2 vế cho ${4^x}$ ta có:

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