Tìm x: 7^x 2 7^x 1 7^x / 57

\[\frac{{{7^{x + 2}} + {7^{x + 1}} + {7^x}}}{{57}} = \frac{{{5^{2x}} + {5^{2x + 1}} + {5^{2x + 3}}}}{{131}}\]

\[\frac{{{7^x} \cdot {7^2} + {7^x} \cdot 7 + {7^x}}}{{57}} = \frac{{{5^{2x}} + {5^{2x}} \cdot 5 + {5^{2x}} \cdot {5^3}}}{{131}}\]

\[\frac{{{7^x}\left( {{7^2} + 7 + 1} \right)}}{{57}} = \frac{{{5^{2x}}\left( {1 + 5 + {5^2}} \right)}}{{131}}\]

\[\frac{{{7^x} \cdot 57}}{{57}} = \frac{{{5^{2x}} \cdot 131}}{{131}}\]

7^x = 5^2x

x = 0

Vậy x = 0.