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=(5/5-5/10+5/10-5/15+.........+5/2015-5/2020)
=(1/5-1/10+1/10-1/20+.......+1/2015-1/2020)
=1/5-1/2020
=403/2020
ai tích mk mk vs
\(\frac{5}{5.10}+\frac{5}{10.15}+.............+\frac{5}{2015.2020}\)
\(=\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+..............+\frac{1}{2015}-\frac{1}{2020}\)
\(=\frac{1}{5}-\frac{1}{2020}\)
\(=\frac{403}{2020}\)
\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)
\(=2.\left(\frac{1}{5.10}+\frac{1}{10.15}+\frac{1}{15.20}+...+\frac{1}{2015.2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
\(=2.\frac{403}{2020}=\frac{403}{1010}\)
\(\frac{2}{5.10}+\frac{2}{10.15}+\frac{2}{15.20}+...+\frac{2}{2015.2020}\)
=\(\frac{2}{5}\left(\frac{5}{5.10}+\frac{5}{10.15}+\frac{5}{15.20}+...+\frac{5}{2015.2020}\right)\)
=\(\frac{2}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+\frac{1}{15}-\frac{1}{20}+...+\frac{1}{2015}-\frac{1}{2016}\right)\)
=\(\frac{2}{5}.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
=\(\frac{2}{5}.\frac{403}{2020}\)
=\(\frac{403}{5005}\)
=(45.5)^10/75^10
=(225/75)^10
=3^10
=59049
\(\frac{45^{10}\cdot5^{10}}{75^{10}}=\frac{3^{10}\cdot15^{10}\cdot5^{10}}{75^{10}}=\frac{3^{10}\cdot3^{10}\cdot5^{10}\cdot5^{10}}{3^{10}\cdot25^{10}}=\frac{3^{10}\cdot3^{10}\cdot5^{10}\cdot5^{10}}{3^{10}\cdot5^{10}\cdot5^{10}}=\frac{3^{20}\cdot5^{20}}{3^{10}\cdot5^{20}}=3^{10}\)
a) |10| + |-10| = 10 + 10 = 20 b) |-9| - |-2| = 9 - 2 = 7
c) |-8| . |-5| = 8. 5 = 40 d) |18| : |-6| = 18 : 6 = 3
Thay x = 10 vào biểu thức ta được:
− 1 . − 2 . − 3 . − 4 − 5 .10 = − 120.10 = − 1200.
Đặt A = \(\frac{10}{5.10}+\frac{10}{10.15}+...+\frac{10}{2015.2020}\)
\(=10\left(\frac{1}{5.10}+\frac{1}{10.15}+...+\frac{1}{2015.2020}\right)\)
\(=\frac{10}{5}\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{2020}\right)\)\(=\frac{2}{5}-\frac{2}{2020}\)
\(=\frac{2}{5}-\frac{1}{1010}\)\(=\frac{404}{1010}-\frac{1}{1010}\)\(=\frac{403}{1010}\)
Vậy giá trị của biểu thức đã cho là 403/1010
\(\frac{10}{5.10}+\frac{10}{10.15}+...+\frac{10}{2015.2020}\)
\(=2.\left(\frac{5}{5.10}+\frac{5}{10.15}+...+\frac{5}{2015.2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{10}+\frac{1}{10}-\frac{1}{15}+...+\frac{1}{2015}-\frac{1}{2020}\right)\)
\(=2.\left(\frac{1}{5}-\frac{1}{2020}\right)\)
\(\frac{2}{5}-\frac{1}{1010}\)
Tính nốt nha