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H
8
30 tháng 6 2015
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}=\frac{2^{10}.3^8-2^{10}.3^8.3}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\frac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8.\left(1+5\right)}=-\frac{2}{6}=-\frac{1}{3}\)
\(B=\frac{4^6.9^5+6^9.120}{8^4.3^{12}-6^{11}}=\frac{2^{12}.3^{10}+2^9.3^9.2^3.3.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{12}.3^{10}+2^{12}.3^{10}.5}{2^{12}.3^{12}-2^{11}.3^{11}}=\frac{2^{11}.2.3^{10}.\left(1+5\right)}{2^{11}.3^{10}.3.\left(6-1\right)}=\frac{12}{15}=\frac{4}{5}\)
30 tháng 6 2015
\(A=\frac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\frac{2^{10}.3^8-2^2.3^9}{2^{10}.3^8-\left(-\left(2^2.3^8.5\right)\right)}=\frac{2^2.3^9}{-\left(2^2.3^8.5\right)}=-\frac{3}{5}\)
NG
2
S
5 tháng 2 2023
\(A=\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}\)
\(=\dfrac{2^{10}.3^8-2.2^9.3^9}{2^{10}.3^8+2^8.3^8.2^2.5}\)
\(=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}\)
\(=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}\)
\(=-\dfrac{2}{6}=-\dfrac{1}{3}\)
NN
1
H9
HT.Phong (9A5)
CTVHS
22 tháng 7 2023
a) \(\dfrac{27^3\cdot11+9^5\cdot5}{3^9\cdot2^4}\)
\(=\dfrac{3^9\cdot11+3^{10}\cdot5}{3^9\cdot2^4}\)
\(=\dfrac{3^9\cdot\left(11+3\cdot5\right)}{3^9\cdot2^4}\)
\(=\dfrac{11+15}{16}\)
\(=\dfrac{26}{16}\)
\(=\dfrac{13}{8}\)
b) \(\dfrac{5^8+2^2\cdot25^4+2^3\cdot125^3-15^4\cdot5^4}{4^2\cdot625^2}\)
\(=\dfrac{5^8+2^2\cdot5^8+2^3\cdot5^9-3^4\cdot5^4\cdot5^4}{2^4\cdot5^8}\)
\(=\dfrac{5^8\cdot\left(1+2^2+2^3\cdot5-3^4\right)}{5^8\cdot2^4}\)
\(=\dfrac{1+4+40-81}{16}\)
\(=\dfrac{-36}{16}\)
\(=\dfrac{-9}{4}\)
c) \(\dfrac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}\)
\(=\dfrac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\)
\(=\dfrac{2^{10}\cdot3^8\cdot\left(1-3\right)}{2^{10}\cdot3^8\cdot\left(1+5\right)}\)
\(=\dfrac{-2}{6}\)
\(=-\dfrac{1}{3}\)
TT
0
IM
22 tháng 7 2016
a)
\(5A=5+5^2+.....+5^{101}\)
\(\Rightarrow5A-A=\left(5+5^2+.....+5^{101}\right)-\left(1+5+.....+5^{100}\right)\)
\(\Rightarrow4A=5^{101}-1\)
\(\Rightarrow A=\frac{5^{101}-1}{4}\)
b)
\(2B=1+\left(\frac{1}{2}\right)^2+....+\left(\frac{1}{2}\right)^{100}\)
\(\Rightarrow2B-B=\left(1+\frac{1}{2^2}+.....+\frac{1}{2^{100}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+......+\frac{1}{2^{99}}\right)\)
\(\Rightarrow B=1-\frac{1}{2^{100}}\)
NN
0
TT
2
8 tháng 9 2017
\(\dfrac{4^5.9^4-2.6^9}{2^{10}.3^8+6^8.20}=\dfrac{2^{10}.3^8-2^{10}.3^9}{2^{10}.3^8+2^{10}.3^8.5}=\dfrac{2^{10}.3^8\left(1-3\right)}{2^{10}.3^8\left(1+5\right)}=-\dfrac{1}{3}\)
19 tháng 3 2021
45.94−2.69210.38+68.20=210.38−210.39210.38+210.38.5=210.38(1−3)210.38(1+5)=−13
DQ
2
27 tháng 2 2020
Ta có: \(\frac{\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}}{\sqrt{\frac{25}{9}}}=\frac{\frac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+3^8\cdot2^8\cdot20}}{\frac{5}{3}}=\frac{2^{10}\left(3^8-3^9\right)}{2^83^8\left(2^2+20\right)}\cdot\frac{3}{5}=\frac{2^83^8\left(1-3\right)\cdot2^2}{2^83^8\cdot24}\cdot\frac{3}{5}\)
\(=\frac{-2\cdot2^2}{24}\cdot\frac{3}{5}=\frac{-8}{24}\cdot\frac{3}{5}=\frac{-24}{120}=\frac{-1}{5}\)
DL
27 tháng 2 2020
\(\frac{4^5\cdot9^4-2\cdot6^9}{2^{10}\cdot3^8+6^8\cdot20}:\sqrt{\frac{25}{9}}\\ =\frac{2^{10}\cdot3^8-2\cdot\left(2\cdot3\right)^9}{2^{10}\cdot3^8+\left(2\cdot3\right)^8\cdot\left(2^2\cdot5\right)}:\frac{5}{3}\\ =\frac{2^{10}\cdot3^8-2\cdot2^9\cdot3^9}{2^{10}\cdot3^8+2^8\cdot3^8\cdot2^2\cdot5}\cdot\frac{3}{5}\\ =\frac{2^{10}\cdot3^8-2^{10}\cdot3^9}{2^{10}\cdot3^8+2^{10}\cdot3^8\cdot5}\cdot\frac{3}{5}\\ =\frac{2^{10}\cdot3^8\left(1-3\right)}{2^{10}\cdot3^8\left(1+5\right)}\cdot\frac{3}{5}\\ =\frac{1-3}{1+5}\cdot\frac{3}{5}\\ =\frac{-2}{6}\cdot\frac{3}{5}=\frac{-1}{5}\)
Đáp án là 1/6 nha
TL
1/6 hok tốt
k cho mk