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

\(4^6.9^5+6^9.120:8^4.3^{12}+6^{11}=\frac{4^6.9^5+6^9.120}{8^4.3^{12}+6^{11}}=\frac{ \left(2^2\right)^6.\left(3^2\right)^5+6^9.6.20}{2^3.3^{12}+6^9.6.6}=\frac{2^{12}.3^{10}+20}{2^3.3^{12}+6}=\frac{2^9+20}{3^2+6}\)

\(\frac{2^9+20}{3^2+6}=\frac{512+20}{9+6}=\frac{532}{15}\)

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

Giải giúp em với mn

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

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\(\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+3^{10}.2^{12}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}.6}{6^{12}-6^{11}}=\frac{2^{12}.3^{10}.6}{6^{11}\left(6-1\right)}=\frac{2^{10}.3^{10}\left(2^2+1\right).6}{6^{11}.5}=\frac{6^{11}.5}{6^{11}.5}=1\)

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

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

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