MN ƠI GIÚP MÌNH VỚI MÌNH CẦN GẤP

\(A=\dfrac{6^3+3\cdot6^2+3^3}{13}\)

\(=\dfrac{3^3\cdot8+3^3\cdot4+3^3}{13}\)

=27

\(=\dfrac{5^3\left(2^3+2+1\right)}{55}-\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)

\(=5^2-\dfrac{2^{12}\cdot3^{10}\cdot\left(1+5\right)}{2^{11}\cdot3^{11}\left(2\cdot3-1\right)}=25-\dfrac{2}{3}\cdot\dfrac{6}{5}\)

=25-4/5

=24,2

a) \(\dfrac{2727-101}{3.303+404}=\dfrac{2626}{909+404}=\dfrac{2626}{1313}=2\)

b) \(\dfrac{8.9-4.15}{12.7-180}=\dfrac{72-60}{84-180}=\dfrac{12}{-96}=\dfrac{-1}{8}\)

c) \(\dfrac{-19}{3^2.7.11}=\dfrac{-19}{9.7.11}=\dfrac{-19}{63.11}=\dfrac{-19}{693}\)

d) \(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\dfrac{2^{12}.3^{10}+120.6^9}{2^{12}.3^{12}-6^{11}}=\dfrac{2^2.6^{10}+20.6.6^9}{6^{12}-6^{11}}=\dfrac{4.6^{10}+20.6^{10}}{6^{11}\left(6-1\right)}=\dfrac{\left(4+20\right).6^{10}}{5.6^{11}}=\dfrac{24}{30}=\dfrac{4}{5}\)

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

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

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

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

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

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

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

E = \(\frac{\left(2^2\right)^6.\left(3^2\right) ^5+\left(2.3\right)^9.2^3.3.5}{-\left(2^3\right)^4.3^{12}-\left(2.3\right)^{11}}\)

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

E = \(\frac{2^{12}.3^{10}+2^{13}.3^{10}.5}{-2^{11}.3^{11}.\left(2.3+1\right)}\)

E = \(\frac{2^{12}.3^{10}.\left(1+5\right)}{-2^{11}.3^{11}.7}\)

E = \(\frac{2^{12}.3^{10}.6}{-2^{11}.3^{11}.7}\)

E=\(\frac{-2^{11}.\left(-2\right).3^{10}.6}{-2^{11}.3^{10}.3.7}\)

E = \(\frac{-2.6}{3.7}=-\frac{4}{7}\)

Vậy E = -4/7 

Ý F bn lm tương tự nha 

thank bn nha