Tính : \(\frac{25^3.3+25^3.5}{2^{27}}\)
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TT
1
30 tháng 4 2018
a) 2,25 : \(\frac{-6}{25}=\frac{9}{4}\cdot\frac{-25}{6}=\frac{9\cdot\left(-25\right)}{4\cdot6}=\frac{3\cdot3\cdot\left(-25\right)}{4\cdot3\cdot2}=\frac{-75}{8}\)
b) \(\frac{2^3\cdot5^2\cdot7^2\cdot3^7}{49\cdot5^3\cdot3^6\cdot11}=\frac{2^3\cdot5^2\cdot7^2\cdot3^6\cdot3}{7^2\cdot5^2\cdot5\cdot3^6\cdot11}=\frac{2^3\cdot3}{5\cdot11}=\frac{24}{55}\)
BC
31 tháng 3 2019
a)Ta có:
\(A=4\frac{25}{16}+25\left(\frac{9}{16}:\frac{125}{64}\right):\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+25.\frac{36}{125}:\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+\frac{36}{5}:\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+\frac{-32}{15}\)
\(\Rightarrow A=\frac{823}{240}\)
Vậy A=.....
b)Ta có:
\(C=\frac{2^3}{3.5}+\frac{2^3}{5.7}+\frac{2^3}{7.9}+...+\frac{2^3}{101.103}\)
\(\Rightarrow C=\frac{2^2.2}{3.5}+\frac{2^2.2}{5.7}+\frac{2^2.2}{7.9}+...+\frac{2^2.2}{101.103}\)
\(\Rightarrow C=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)
\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{101}-\frac{1}{103}\right)\)
\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{103}\right)\)
\(\Rightarrow C=4.\frac{100}{309}\)
\(\Rightarrow C=\frac{400}{309}\)
Vậy C=.....
ND
Nguyễn Đức Trí
VIP
17 tháng 8 2023
a) \(...=2^{14}:\left(2^6.2^5\right)=2^{14}:2^{11}=2^3=8\)
b) \(...=5^{25}.5^2:5^{24}=5^{27}:5^{24}=5^3=125\)
c) \(...=2^{22}:2^2-2^{20}=2^{20}-2^{20}=0\)
d) \(...=3^6:3^3=3^3=27\)
13 tháng 7 2015
\(\frac{25^4.16^3.3^7}{75^4.6^3.8^2}=\frac{5^8.4^6.3^7}{3^4.25^4.2^3.3^3.2^6}=\frac{3^7.2^{12}.5^8}{3^7.5^8.2^9}=2^3=8\)
AY
20 tháng 9 2016
\(a.27.75+25.27-150\)
\(=27.\left(75+25\right)-150\)
\(=27.100-150\)
\(=2700-150\)
\(=2550\)
16 tháng 9 2018
a, 27.75+25.27-150 b,3.52-16:23 d, 60 : {[( 12-3 ).2] +2}
= 27.(75+25)-150 = 3.25-16:4 = 60 : {[ 9.2] + 2
= 27. 100 - 150 = 75 - 4 = 60 : 18 + 2
= 2700 - 150 = 71 = 60 : 20
= 2550 = 3
Tk mik nha
BN
12 tháng 2 2017
\(\frac{2}{3.5}+\frac{3}{5.8}+\frac{11}{8.19}+\frac{13}{19.32}+\frac{25}{32.57}+\frac{30}{57.87}\)
\(=\frac{5-3}{3.5}+\frac{8-5}{3}+\frac{19-8}{8.19}+\frac{32-29}{19.32}+\frac{57-32}{32.57}+\frac{87-57}{57.87}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\frac{1}{8}-\frac{1}{19}+\frac{1}{19}-\frac{1}{32}+\frac{1}{32}-\frac{1}{57}+\frac{1}{57}-\frac{1}{87}\)
\(=\frac{1}{3}-\frac{1}{87}=\frac{28}{87}\)
23 tháng 4 2019
\(\frac{2^35^27^23^7}{49.5^3.3^6.11}\)
= \(\frac{2^31^21^21^7}{7.1^3.1^6.11}\)
= \(\frac{8}{77}\)
Đặt \(A=\frac{25^3.3+25^3.5}{2^{27}}\)
Ta có:\(A=\frac{25^3\left(3+5\right)}{2^{27}}\)
\(A=\frac{25^3.8}{2^{27}}\)
\(A=\frac{25^3.2^3}{\left(2^9\right)^3}\)
\(A=\frac{50^3}{512^3}\)
\(A=\left(50:512\right)^3\)
\(A=\left(\frac{25}{256}\right)^3\)
Vậy \(A=\left(\frac{25}{256}\right)^3\)
\(\frac{25^3\cdot3+25^3\cdot5}{2^{27}}=\frac{25^3\left(3+5\right)}{2^{27}}=\frac{5^3\cdot2^3}{2^{27}}=\frac{10^3}{\left(2^9\right)^3}=\left(\frac{5}{256}\right)^3\)