Ta có : 

\(A=\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\frac{2^{12}.3^{10}+2^3.3.5.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}=\frac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(6+1\right)}=\frac{12}{21}=\frac{4}{7}\)

Chúc bạn học tốt ~

\(A=\dfrac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}=\dfrac{\left(4^2\right)^3.3^{10}+2^3.3.5\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)

\(A=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}=\dfrac{2^{11}.3^{10}\left(2+2.5\right)}{2^{11}.3^{10}\left(2.3^2+3\right)}\)

\(A=\dfrac{2+10}{18+3}=\dfrac{12}{21}=\dfrac{4}{7}\)

Ta có: \(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)

\(=\frac{\left(2^4\right)^3.3^{10}+2^3.3.5.\left(2.3\right)^9}{\left(2^2\right)^6.3^{12}+\left(2.3\right)^{11}}\)

\(=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}+2^{11}.3^{11}}\)

\(=\frac{2^{12}.3^{10}.\left(1+5\right)}{2^{11}.3^{11}\left(6+1\right)}\)

\(=\frac{2.6}{3.7}=\frac{4}{7}\)

\(\frac{16^3.3^{10}+120.6^9}{4^6.3^{12}+6^{11}}\)

\(=\frac{2^{12}.3^{10}+120.2^9.3^9}{2^{12}.3^{12}+2^{11}.3^{11}}\)

\(=\frac{2^3.2^9.3^9.3+120.2^9.3^9}{2.2^{11}.3^{11}.3+2^{11}.3^{11}}\)

\(=\frac{2^9.3^9.\left(2^3.3+120\right)}{2^{11}.3^{11}.\left(2.3+1\right)}\)

\(=\frac{2^9.3^9.144}{2^{11}.3^{11}.7}\)

\(=\frac{2^9.3^9.2^4.3^2}{2^{11}.3^{11}.7}\)

\(=\frac{2^2.2^7.3^9.2^4.3^2}{2^{11}.3^{11}.7}\)

\(=\frac{4.2^{11}.3^{11}}{2^{11}.3^{11}.7}=\frac{4}{7}\)

a) \(3^{30}\) và \(5^{20}.\)

Ta có:

\(3^{30}=\left(3^3\right)^{10}=27^{10}.\)

\(5^{20}=\left(5^2\right)^{10}=25^{10}.\)

Vì \(27>25\) nên \(27^{10}>25^{10}.\)

\(\Rightarrow3^{30}>5^{20}.\)

Chúc bạn học tốt!

a/ Có: \(3^{30}=\left(3^3\right)^{10}=27^{10}\\ 5^{20}=\left(5^2\right)^{10}=25^{10}\)

Mà \(27^{10}>25^{10}\)

\(\Rightarrow3^{30}=5^{20}\)

b/ \(A=\frac{16^3\cdot3^{10}+120\cdot6^9}{4^6\cdot3^{12}+6^{11}}\\ A=\frac{\left(2^4\right)^3\cdot3^{10}+2^3\cdot3\cdot5\cdot2^9\cdot3^9}{\left(2^2\right)^6\cdot3^{12}+2^{11}\cdot3^{11}}\\ A=\frac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\\ A=\frac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3+1\right)}\\ A=\frac{2\cdot6}{3\cdot7}=\frac{4}{7}\)

\(\frac{3^{17}\cdot81^{11}}{27^{10}\cdot9^{15}}\)

\(=\frac{3^{17}\cdot\left(3^4\right)^{11}}{\left(3^3\right)^{10}\cdot\left(3^2\right)^{15}}\)

\(=\frac{3^{17}\cdot3^{44}}{3^{30}\cdot3^{30}}\)

\(=\frac{3^{61}}{3^{60}}\)

\(=3\)

\(\frac{9^2\cdot2^{11}}{16^2\cdot6^3}\)

\(=\frac{\left(3^2\right)^2\cdot2^{11}}{\left(2^4\right)^2\cdot\left(2\cdot3\right)^3}\)

\(=\frac{3^4\cdot2^{11}}{2^8\cdot2^3\cdot3^3}\)

\(=\frac{3^4\cdot2^{11}}{2^{11}\cdot3^3}\)

\(=\frac{3^4}{3^3}\)

\(=3\)

\(\frac{2^{10}\cdot3^{31}+2^{40}\cdot3^6}{2^{11}\cdot3^{31}+2^{41}\cdot3^6}\)

\(=\frac{2^{10}\cdot3^{31}+2^{40}\cdot3^6}{2\cdot\left(2^{10}\cdot3^{31}+2^{40}\cdot3^6\right)}\)

\(=\frac{1}{2}\)

\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-2}{6}=-\frac{1}{3}\)

Ta có:\

\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)

\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)

\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(A=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)

\(A=-\frac{2}{6}=-\frac{1}{3}\)

Ta có : 

\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)

\(A=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.4.5}\)

\(A=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)

\(A=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)

\(A=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)

\(A=\frac{1-3}{1+5}\)

\(A=\frac{-2}{6}\)

\(A=\frac{-1}{3}\)

Chúc bạn học tốt ~