A=\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}\)
Chứng tỏ rằng \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+......+\dfrac{1}{99.100}=\dfrac{1}{51}+\dfrac{1}{52}+......+\dfrac{1}{100}\)
cho A=\(\dfrac{1}{1.2}+\dfrac{1}{3.4}+\dfrac{1}{5.6}+...+\dfrac{1}{2013.2014}\)
B=\(\dfrac{1}{1008.2014}+\dfrac{1}{1009.2013}+\dfrac{1}{1010.2012}+...+\dfrac{1}{2014.1008}\)
CTR:\(\dfrac{A}{B}\) là số nguyên
Cho hai biểu thức M = \(\dfrac{1}{1.2}+\dfrac{1}{3.4}+...+\dfrac{1}{37.38}\) và N = \(\dfrac{1}{20.38}+\dfrac{1}{21.37}+...+\dfrac{1}{38.20}\), chứng minh rằng \(\dfrac{M}{N}\) là một số nguyên.
Help me!
\(a.\left(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{9.10}\right).100-\left[\dfrac{5}{2}:\left(x+\dfrac{206}{100}\right)\right]:\dfrac{1}{2}=89\)
1) Rút gọn
A =\(\dfrac{\dfrac{1}{19}+\dfrac{1}{18}+\dfrac{1}{17}+.......+\dfrac{18}{2}+\dfrac{19}{1}}{\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+.......+\dfrac{1}{19}+\dfrac{1}{20}}\)
2) Tìm x
a/ \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+.......+\dfrac{1}{x.\left(x+1\right)}=\dfrac{2016}{2017}\)
Thực hiện phép tính sau:
1, B = \(\dfrac{1}{2}\)+\(\dfrac{1}{2^2}\)+\(\dfrac{1}{2^3}\)+\(\dfrac{1}{2^4}\)+..........+\(\dfrac{1}{2^{2017}}\)
2,\(\dfrac{12}{1.2}\).\(\dfrac{2^2}{2.3}\).\(\dfrac{3^2}{3.4}\).\(\dfrac{4^2}{4.5}\).\(\dfrac{5^2}{5.6}\)
Giúp mình nhé dấu (.) là nhân nha
TÍnh A=\(\dfrac{1}{1.2}-\dfrac{1}{1.2.3}+\dfrac{1}{2.3}-\dfrac{1}{2.3.4}+\dfrac{1}{3.4}-\dfrac{1}{3.4.5}+...+\dfrac{1}{99.100}-\dfrac{1}{99.100.101}\)
B=\(\dfrac{5}{1.2.3.4}+\dfrac{5}{2.3.4.5}+...+\dfrac{5}{98.99.100.101}\)
C=\(\dfrac{6}{1^2+2^2}+\dfrac{10}{2^2+3^2}+\dfrac{14}{3^2+4^2}+...+\dfrac{398}{99^2.100^2}\)