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27 tháng 7 2021
a) Ta có: \(\dfrac{-5}{7}\left(\dfrac{14}{5}-\dfrac{7}{10}\right):\left|-\dfrac{2}{3}\right|-\dfrac{3}{4}\left(\dfrac{8}{9}+\dfrac{16}{3}\right)+\dfrac{10}{3}\left(\dfrac{1}{3}+\dfrac{1}{5}\right)\)
\(=\dfrac{-5}{7}\cdot\dfrac{3}{2}\cdot\dfrac{21}{10}-\dfrac{3}{4}\cdot\dfrac{56}{3}+\dfrac{10}{3}\cdot\dfrac{8}{15}\)
\(=\dfrac{-9}{4}-14+\dfrac{16}{9}\)
\(=\dfrac{-1621}{126}\)
b) Ta có: \(\dfrac{17}{-26}\cdot\left(\dfrac{1}{6}-\dfrac{5}{3}\right):\dfrac{17}{13}-\dfrac{20}{3}\left(\dfrac{2}{5}-\dfrac{1}{4}\right)+\dfrac{2}{3}\left(\dfrac{6}{5}-\dfrac{9}{2}\right)\)
\(=\dfrac{-17}{26}\cdot\dfrac{13}{17}\cdot\dfrac{-3}{2}-\dfrac{20}{3}\cdot\dfrac{3}{20}+\dfrac{2}{3}\cdot\dfrac{-33}{10}\)
\(=\dfrac{3}{4}-1-\dfrac{11}{5}\)
\(=-\dfrac{49}{20}\)
24 tháng 9 2018
Có : \(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\dfrac{1-3}{1+5}=\dfrac{1}{3}\)
NT
2
TT
15 tháng 11 2021
\(\dfrac{4^{^5}.9^{^4}-2.6^{^9}}{2^{^{10}}.3^{^8}+6^{^8}.20}\)=\(\dfrac{2^{^{10}}.3^{^8}-2.2^{^9}.3^{^9}}{2^{^{10}}.3^{^8}+2^{^8}.3^{^8}.2^{^2}.5}\)=\(\dfrac{2^{^{10}}.3^{^8}-2^{^{10}}.3^{^9}}{2^{^{10}}.3^{^8}+2^{^{10}}.3^{^8}.5}\)=\(\dfrac{2^{^{10}}.3^{^8}\left(1-3\right)}{2^{^{10}}.3^{^8}\left(1+5\right)}\)=\(\dfrac{1-3}{1+5}\)=\(\dfrac{1}{3}\)
26 tháng 11 2021
45.94−2.69210.38+68.2045.94−2.69210.38+68.20=210.38−2.29.39210.38+28.38.22.5210.38−2.29.39210.38+28.38.22.5=210.38−210.39210.38+210.38.5210.38−210.39210.38+210.38.5=210.38(1−3)210.38(1+5)210.38(1−3)210.38(1+5)=1−31+51−31+5=13
9 tháng 7 2021
Ta có: \(\dfrac{5\cdot4^{15}\cdot9^9-4\cdot3^{20}\cdot8^9}{5\cdot2^{10}\cdot6^{19}-7\cdot2^{29}\cdot27^6}\)
\(=\dfrac{5\cdot2^{30}\cdot3^{18}-2^2\cdot2^{27}\cdot3^{20}}{5\cdot2^{10}\cdot2^{19}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{3^{18}\cdot2^{29}\left(5\cdot2-3^2\right)}{5\cdot2^{29}\cdot3^{19}-7\cdot2^{29}\cdot3^{18}}\)
\(=\dfrac{3^{18}\cdot2^{29}}{2^{29}\cdot3^{18}\left(5\cdot2-7\right)}\)
\(=\dfrac{1}{3}\)
HT
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