g) Ta có
$\dfrac{4^6 . 9^5 + 6^9 . 120}{8^4 . 3^{12} - 6^{11}} = \dfrac{(2^2)^6 . (3^2)^5 + (2.3)^9 . 2^3.3.5}{(2^3)^4 . 3^{12} - (2.3)^{11}}$
$= \dfrac{2^{12} . 3^{10} + 2^{12} . 3^{10} . 5}{2^{12} . 3^{12} - 2^{11} . 3^{11}}$
$= \dfrac{2^{11} . 3^{10}(2 + 2.5)}{2^{11} . 3^{11} (2.3 - 1)}$
$= \dfrac{12}{3.5} = \dfrac{4}{5}$