Tính:
a) \(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
b) T=\(\dfrac{5^{16}.27^7}{125^5.9^{11}}\)
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28 tháng 10 2021
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1-5\right)}=\dfrac{-2}{-4}=\dfrac{1}{2}\)
20 tháng 9 2021
\(A=\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^{10}\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)
\(=-\dfrac{2}{6}=-\dfrac{1}{3}\)
4 tháng 1 2023
\(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot6}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
26 tháng 8 2017
\(B=\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.\left(6\right)^8.6}{2^{10}.3^8+\left(6\right)^8.20}=\dfrac{2^{10}.3^8-2\left(3.2\right)^8.6}{2^{10}.3^8+\left(3.2\right)^8.20}\)
\(B=\dfrac{2^{10}.3^8-2.3^8.2^8.6}{2^{10}.3^8+3^8.2^8.20}=\dfrac{3^8\left(2^{10}-2.2^8.6\right)}{3^8\left(2^{10}+2^8.20\right)}=\dfrac{2^{10}-2^9.6}{2^{10}+2^8.20}\)
\(B=\dfrac{2^8\left(2^2-2.6\right)}{2^8\left(2^2+20\right)}=\dfrac{2^2-2.6}{2^2+20}=\dfrac{4-12}{4+20}=\dfrac{-8}{24}=\dfrac{-1}{3}\)
NN
1
19 tháng 12 2017
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{\left(2^2\right)^5.\left(3^2\right)^4-2.6^8.6}{2^{10}.3^8+6^8.20}\)
\(=\dfrac{2^{10}.3^8-2.6^8.6}{2^{10}.3^8+6^8.2.2.5}=\dfrac{-1}{3}\)
NN
1
LD
19 tháng 12 2017
\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}=\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot\left(2\cdot3\right)^9}{2^{10}\cdot3^8+\left(2\cdot3\right)^8\cdot4\cdot5}=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+1\right)}=\dfrac{1-3}{1+1}=-\dfrac{2}{2}=-1\)
NN
1
MV
1 tháng 11 2017
\(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\\ =\dfrac{\left(2^2\right)^5\cdot\left(3^2\right)^4-2\cdot\left(2\cdot3\right)^9}{2^{10}\cdot3^8+\left(2\cdot3\right)^8\cdot20}\\ =\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot20}\\ =\dfrac{2^8\cdot3^8\left(2^2-2^2\cdot3\right)}{2^8\cdot3^8\left(2^2+20\right)}\\ =\dfrac{4-4\cdot3}{4+20}\\ =\dfrac{-8}{24}\\ =\dfrac{-1}{3}\)
a: \(=\dfrac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}=\dfrac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}=\dfrac{-2}{6}=\dfrac{-1}{3}\)
b: \(=\dfrac{5^{16}\cdot3^{21}}{5^{15}\cdot3^{22}}=\dfrac{5}{3}\)