\(=\dfrac{-2^{12}\cdot3^{10}-2^{12}\cdot3^{10}\cdot5}{-2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2\cdot6}{3\cdot7}=\dfrac{12}{21}=\dfrac{4}{7}\)

\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}\)

\(=\dfrac{2}{3}\cdot\dfrac{6}{5}=\dfrac{12}{15}=\dfrac{4}{5}\)

\(\dfrac{4^6.9^5+6^9.120}{8^4.3^2-6^{11}}\)

\(=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.120}{\left(2^3\right)^4.3^2-\left(2.3\right)^{11}}\)

\(=\dfrac{2^{12}.3^{10}+2^9.3^9.120}{2^{12}.3^2-2^{11}.3^{11}}\)

\(=\dfrac{1.3^8+1.1.120}{1.1-4.9}\)

\(=\dfrac{1.6561+120}{1-36}=\dfrac{6681}{-35}\)

Chắc sai đấy!!!!

\(\dfrac{4^6.9^5+6^9.120}{8^4.3^2-6^{11}}\)

\(=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+\left(2.3\right)^9.\left(2^3.3.5\right)}{\left(2^3\right)^4.3^2-\left(2.3\right)^{11}}\)

\(=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^2-2^{11}.3^{11}}\)

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

\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^2\left(2-3^9\right)}\)

\(=\dfrac{2.3^8.6}{2-3^9}=\dfrac{2.3^8.2.3}{2-3^9}=\dfrac{2^2.3^9}{2-3^9}\)

GIÚP MÌNH VỚI MN ƠI

\(C=\dfrac{6^3+3\cdot6^2+3^3}{13}=\dfrac{3^3\cdot8+3^3\cdot4+3^3}{13}=27\)

\(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}\)
\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.5}\)
\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}=\dfrac{2.6}{3.5}=\dfrac{4}{5}\)

Giải giúp em với mn

\(A=\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{\left(2^2\right)^6.\left(3^2\right)^5+6^9.6.2^2.5}{\left(2^3\right)^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\dfrac{2^{11}.3^{10}\left(2^1+2^1.5\right)}{2^{11}.3^{10}\left(2^1.3^2-1.3^1\right)}=\dfrac{2+10}{2.9-1.3}=\dfrac{12}{15}=\dfrac{4}{5}\)

\(=\dfrac{2^{12}\cdot3^{10}+2^9\cdot2^3\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot5}\)

\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot5}=\dfrac{2\cdot6}{3\cdot5}=\dfrac{12}{15}=\dfrac{4}{5}\)

ta có : \(\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\dfrac{4\left(4^5.9^5+5.6^{10}\right)}{-2^{12}.3^{12}-6^{11}}=\dfrac{4\left(2^{10}.3^{10}+5.6^{10}\right)}{-2^{12}.3^{12}-6^{11}}\)

\(=\dfrac{4\left(6^{10}+5.6^{10}\right)}{-6^{12}-6^{11}}=\dfrac{4.6^{11}}{-6^{11}\left(6+1\right)}=-\dfrac{4}{7}\)

\(=\dfrac{4}{5}\)

\(=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{-2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)

\(=-\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}=\dfrac{-2}{3}\)

Giải:

\(10.\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)

\(=10.\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)

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

\(=10.\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}\left(2.3-1\right)}\)

\(=10.\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}\)

\(=10.\dfrac{2^{13}.3^{11}}{2^{11}.3^{11}.5}\)

\(=10.\dfrac{2^2}{5}\)

\(=2^3=8\)

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