Tính A=\(\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}\)
K
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21 tháng 6 2023
\(=\dfrac{-2^{12}\cdot3^{10}-2^{12}\cdot3^{10}\cdot5}{-2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot7}=\dfrac{2\cdot6}{3\cdot7}=\dfrac{12}{21}=\dfrac{4}{7}\)
L
4
S
5 tháng 1 2023
\(\dfrac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}\)
\(=\dfrac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{11}.3^{11}\left(2.3-1\right)}\)
\(=\dfrac{2^{12}.3^{10}\left(1+5\right)}{2^{11}.3^{11}.5}\)
\(=\dfrac{2^{12}.3^{10}.6}{2^{11}.3^{11}.5}=\dfrac{2.6}{3.5}=\dfrac{4}{5}\)
27 tháng 5 2022
\(=\dfrac{2^{12}\cdot3^{10}+2^9\cdot2^3\cdot3^9\cdot3\cdot5}{2^{12}\cdot3^{12}-2^{11}\cdot3^{11}}\)
\(=\dfrac{2^{12}\cdot3^{10}+2^{12}\cdot3^{10}\cdot5}{2^{11}\cdot3^{11}\cdot5}\)
\(=\dfrac{2^{12}\cdot3^{10}\cdot6}{2^{11}\cdot3^{11}\cdot5}=\dfrac{2\cdot6}{3\cdot5}=\dfrac{12}{15}=\dfrac{4}{5}\)
3 tháng 10 2018
ta có : \(\dfrac{4^6.9^5+6^9.120}{-8^4.3^{12}-6^{11}}=\dfrac{4\left(4^5.9^5+5.6^{10}\right)}{-2^{12}.3^{12}-6^{11}}=\dfrac{4\left(2^{10}.3^{10}+5.6^{10}\right)}{-2^{12}.3^{12}-6^{11}}\)
\(=\dfrac{4\left(6^{10}+5.6^{10}\right)}{-6^{12}-6^{11}}=\dfrac{4.6^{11}}{-6^{11}\left(6+1\right)}=-\dfrac{4}{7}\)
12 tháng 6 2022
\(=\dfrac{2^{12}\cdot3^{10}+2^9\cdot3^9\cdot2^3\cdot3\cdot5}{-2^{12}\cdot3^{12}+2^{11}\cdot3^{11}}\)
\(=-\dfrac{2^{12}\cdot3^{10}\left(1+5\right)}{2^{11}\cdot3^{11}\cdot\left(2\cdot3-1\right)}=\dfrac{-2}{3}\)
TH
2
3 tháng 10 2021
\(C=\dfrac{6^3+3\cdot6^2+3^3}{13}=\dfrac{3^3\cdot8+3^3\cdot4+3^3}{13}=27\)
HN
1
HH
24 tháng 6 2018
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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