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19 tháng 3 2021
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}:\sqrt{\dfrac{25}{9}}=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}:\dfrac{5}{3}\)
\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}.\dfrac{5}{3}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1-5\right)}.\dfrac{5}{3}=\dfrac{1-3}{1-5}.\dfrac{5}{3}=\dfrac{1}{2}.\dfrac{5}{3}=\dfrac{5}{6}\)
19 tháng 3 2021
\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\div\sqrt{\dfrac{25}{9}}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\div\dfrac{5}{4}\)
=\(\dfrac{2^{10}\cdot3^8\left(1-2\cdot3\right)}{2^{10}\cdot3^8\left(1+5\right)}\div\dfrac{5}{4}\)
=\(\dfrac{1-6}{1+5}\cdot\dfrac{4}{5}\)
=\(-\dfrac{5}{6}\cdot\dfrac{4}{5}\)
=\(-\dfrac{2}{3}\)
ND
27 tháng 2 2018
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)
\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)
\(=\dfrac{-1}{3}\)
20 tháng 9 2021
\(A=\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^{10}\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)
\(=-\dfrac{2}{6}=-\dfrac{1}{3}\)
24 tháng 1 2017
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-2}{6}=\frac{-1}{3}\)
TE
22 tháng 6 2015
\(4^5\cdot9^4-2.6^9=\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9=2^{10}.3^8-2.2^9.3^9=2^{10}.3^8-2^{10}.3^9=3^8-3^9=-13122\)
\(2^{10}.3^8+6^8.20=2^{10}.3^8+\left(2.3\right)^8.2^2.5=2^{10}.3^8+2^8.3^8.2^2.5=2^{10}.3^8+2^{10}.3^8.5=2^{10}.3^8.\left(1+5\right)=2^{10}.3^8.6=2^{10}.3^8.2.3=2^{11}.3^9\)
D
20 tháng 4 2015
\(\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.2^2.5}\)\(=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=\frac{-2}{6}=\frac{-1}{3}\)
9 tháng 1 2017
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(=\frac{2^{10}.3^8-2.\left(2.3\right)^9}{2^{10}.3^8+\left(2.3\right)^8.\left(2^2.5\right)}\)
\(=\frac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)
\(=\frac{-2}{6}\)
\(=\frac{-1}{3}\)
Vậy \(A=\frac{-1}{3}\)