Ta có: \(\frac{\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}}{\sqrt{\frac{25}{9}}}=\frac{\frac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^8\cdot20}}{\frac{5}{3}}=\frac{2^{10}\left(3^8-3^9\right)}{2^83^8\left(2^2+20\right)}\cdot\frac{3}{5}=\frac{2^83^8\left(1-3\right)\cdot2^2}{2^83^8\cdot24}\cdot\frac{3}{5}\)
\(=\frac{-2\cdot2^2}{24}\cdot\frac{3}{5}=\frac{-8}{24}\cdot\frac{3}{5}=\frac{-24}{120}=\frac{-1}{5}\)
\(\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}:\sqrt{\frac{25}{9}}\\ =\frac{2^{10}\cdot3^8-2\cdot\left(2\cdot3\right)^9}{2^{10}\cdot3^8+\left(2\cdot3\right)^8\cdot\left(2^2\cdot5\right)}:\frac{5}{3}\\ =\frac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\cdot\frac{3}{5}\\ =\frac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\cdot\frac{3}{5}\\ =\frac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}\cdot\frac{3}{5}\\ =\frac{1-3}{1+5}\cdot\frac{3}{5}\\ =\frac{-2}{6}\cdot\frac{3}{5}=\frac{-1}{5}\)