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HN
1
8 tháng 8 2015
\(\frac{2048^2.2^{11}.8^6.64^7.128^3}{1024^4.4^8.16^7.2^{30}}=\frac{2^{22}.2^{11}.2^{18}.2^{42}.2^{21}}{2^{40}.2^{16}.2^{28}.2^{30}}=\frac{2^{114}}{2^{114}}=1\)
QN
12 tháng 7 2017
\(F=\frac{\left(\frac{2}{5}\right)^7.5^7+\left(\frac{9}{4}\right)^9\div\left(\frac{3}{16}\right)^3}{2^7.5^2+512}\)
\(F=\frac{\left(\frac{2.5}{5}\right)^7+\left(\frac{9.16}{4.3}\right)^3}{2^7.5^2+2^9}=\frac{2^7+12^3}{2^7.5^2+2^9}=\frac{2^7+2^6.3^3}{2^7.5^2+2^9}=\frac{2^6.\left(2+3^3\right)}{2^7.\left(5^2+2^2\right)}=\frac{2^6.29}{2^7.29}\)
\(F=\frac{1}{2}\)
S
19 tháng 6 2019
\(2A=1+\frac{1}{2}+\frac{1}{4}+....+\frac{1}{512}\Rightarrow2A-A=1-\frac{1}{1024}=\frac{1023}{1024}\)
19 tháng 6 2019
\(A=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\)
\(2A=1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\)
\(2A-A=\left[1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{512}\right]-\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}\right]\)
\(A=1-\frac{1}{2014}=\frac{2013}{2014}\)
TT
3
XO
2 tháng 2 2020
Ta có : \(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{512}+\frac{1}{1024}=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)
Đặ A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\)(1)
=> 2A = \(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\)(2)
Lấy (2) trừ (1) theo vế ta có :
2A - A = \(\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^8}+\frac{1}{2^9}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^9}+\frac{1}{2^{10}}\right)\)
=> A = \(1-\frac{1}{2^{10}}=\frac{2^{10}-1}{2^{20}}\)
2 tháng 2 2020
\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{10}}\)
\(\Leftrightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^9}\)
\(\Rightarrow2A-A=1-\frac{1}{2^{10}}=\frac{1023}{1024}\)
PA
10 tháng 7 2016
a.
\(\frac{5^3\times90\times4^3}{25^2\times3^2\times2^{13}}=\frac{5^3\times5\times9\times2\times\left(2^2\right)^3}{\left(5^2\right)^2\times3^2\times2^{13}}=\frac{5^4\times3^2\times2\times2^6}{5^4\times3^2\times2^{13}}=\frac{1}{2^6}=\frac{1}{64}\)
b.
\(\frac{18\times27+18\times\left(-23\right)}{34\times4-4\times52}=\frac{18\times\left(27-23\right)}{4\times\left(34-52\right)}=\frac{18\times4}{4\times\left(-18\right)}=-1\)
c.
\(\frac{15^2\times16^4-15^3\times16^3}{12^2\times20^3-20^2\times12^3}=\frac{16^3\times15^2\times\left(16-15\right)}{12^2\times20^2\times\left(20-12\right)}=\frac{16\times\left(16\times15\right)^2}{8\times\left(20\times12\right)^2}=\frac{16\times240^2}{8\times240^2}=2\)
d.
\(\frac{2\times3+4\times6+14\times21}{3\times5+6\times10+21\times35}=\frac{2\times3\times\left(1+2\times2+7\times7\right)}{3\times5\times\left(1+2\times2+7\times7\right)}=\frac{2}{5}\)
Chúc bạn học tốt
BY
10 tháng 7 2016
a) \(\frac{5^3\cdot90\cdot4^3}{25^2\cdot3^2\cdot2^{13}}=\frac{5^3\cdot2\cdot3^2\cdot5\cdot2^6}{5^4\cdot3^2\cdot2^{13}}=\frac{1}{2^6}=\frac{1}{64}\)
b) \(\frac{18\cdot27+18\cdot\left(-23\right)}{34\cdot4-4\cdot52}=\frac{18\left(27-23\right)}{4\left(34-52\right)}=\frac{9\cdot4}{2\cdot\left(-18\right)}=\frac{3^2\cdot2^2}{2\cdot2\cdot3^2\cdot\left(-1\right)}=-1\)
c) \(\frac{15^2\cdot16^4-15^3\cdot16^3}{12^2\cdot20^3-20^2\cdot12^3}=\frac{15^2\cdot16^3\left(16-15\right)}{12^2\cdot20^2\left(20-12\right)}=\frac{15^2\cdot16^3}{12^2\cdot20^2\cdot8}=\frac{3^2\cdot5^2\cdot2^{12}}{2^4\cdot3^2\cdot2^4\cdot5^2\cdot2^3}=2\)
d) \(\frac{2\cdot3+4\cdot6+14\cdot21}{3\cdot5+6\cdot10+21\cdot35}=\frac{2\cdot3+2^2\cdot2\cdot3+2\cdot3\cdot7^2}{3\cdot5+2^2\cdot3\cdot5+3\cdot5\cdot7^2}=\frac{2\cdot3\left(1+2^2+7^2\right)}{3\cdot5\left(1+2^2+7^2\right)}=\frac{2}{5}\)
VP
0
VP
0
\(\frac{512^4.128^6.64^2.2^{13}.4^7}{1024^2.64^5.128^3.8^4.512^4}=\frac{2^{117}}{2^{119}}=\frac{2^{117}}{2^{117}.2^2}=\frac{1}{2^2}=\frac{1}{4}\)